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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 301. |
Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are con – cyclic. |
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Answer» Given: The points (3, -2), (1, 0), (-1, -2) and (1, -4) Let us assume the circle passes through the points A, B, C. So by using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1) Substitute the points A (3, – 2) in equation (1), we get, 32 + (-2)2 + 2a(3) + 2b(-2) + c = 0 9 + 4 + 6a – 4b + c = 0 6a – 4b + c + 13 = 0….. (2) Substitute the points B (1, 0) in equation (1), we get, 12 + 02 + 2a(1) + 2b(0) + c = 0 1 + 2a + c = 0 ……- (3) Substitute the points C (-1, -2) in equation (1), we get, (- 1)2 + (- 2)2 + 2a(- 1) + 2b(- 2) + c = 0 1 + 4 – 2a – 4b + c = 0 5 – 2a – 4b + c = 0 2a + 4b – c – 5 = 0….. (4) Upon simplifying the equations (2), (3) and (4) we get, a = – 1, b = 2 and c = 1 Substituting the values of a, b, c in equation (1), we get x2 + y2 + 2(- 1)x + 2(2)y + 1 = 0 x2 + y2 – 2x + 4y + 1 = 0 ….. (5) Now by substituting the point D (1, -4) in equation (5) we get, 12 + (- 4)2 – 2(1) + 4(- 4) + 1 1 + 16 – 2 – 16 + 1 0 ∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) are con – cyclic. |
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| 302. |
Show that the points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle, and find its equation, centre, and radius. |
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Answer» Given: The points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle. Let us assume the circle passes through the points A, B, C. So by using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1) Substituting A (5, 5) in (1), we get, 52 + 52 + 2a(5) + 2b(5) + c = 0 25 + 25 + 10a + 10b + c = 0 10a + 10b + c + 50 = 0….. (2) Substitute the points B (6, 4) in equation (1), we get, 62 + 42 + 2a(6) + 2b(4) + c = 0 36 + 16 + 12a + 8b + c = 0 12a + 8b + c + 52 = 0….. (3) Substitute the point C (-2, 4) in equation (1), we get, (-2)2 + 42 + 2a(-2) + 2b(4) + c = 0 4 + 16 – 4a + 8b + c = 0 20 – 4a + 8b + c = 0 4a – 8b – c – 20 = 0….. (4) Upon simplifying equations (2), (3) and (4) we get, a = – 2, b = – 1 and c = – 20 Now by substituting the values of a, b, c in equation (1), we get x2 + y2 + 2(- 2)x + 2(- 1)y – 20 = 0 x2 + y2 – 4x – 2y – 20 = 0 ….. (5) Substituting D (7, 1) in equation (5) we get, 72 + 12 – 4(7) – 2(1) – 20 49 + 1 – 28 – 2 – 20 0 ∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) lie on a circle. Now let us find the centre and the radius. Centre = (-a, -b) Radius = √(a2 + b2 – c) Comparing equation (5) with equation (1), we get Centre = [-(-4)/2, -(-2)/2)] = (2, 1) Radius = √(22 + 12 – (-20)) = √(25) = 5 ∴ The centre and radius of the circle is (2, 1) and 5. |
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| 303. |
Find the equation of the circle which circumscribes the triangle formed by the lines:(i) x + y + 3 = 0, x – y + 1 = 0 and x = 3(ii) 2x + y – 3 = 0, x + y – 1 = 0 and 3x + 2y – 5 = 0(iii) x + y = 2, 3x – 4y = 6 and x – y = 0(iv) y = x + 2, 3y = 4x and 2y = 3x |
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Answer» (i) x + y + 3 = 0, x – y + 1 = 0 and x = 3 Given: The lines x + y + 3 = 0 x – y + 1 = 0 x = 3 On solving these lines we get the intersection points A (-2, -1), B (3, 4), C (3, -6) So by using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1) Substitute the points (-2, -1) in equation (1), we get (- 2)2 + (- 1)2 + 2a(-2) + 2b(-1) + c = 0 4 + 1 – 4a – 2b + c = 0 5 – 4a – 2b + c = 0 4a + 2b – c – 5 = 0….. (2) Substitute the points (3, 4) in equation (1), we get 32 + 42 + 2a(3) + 2b(4) + c = 0 9 + 16 + 6a + 8b + c = 0 6a + 8b + c + 25 = 0….. (3) Substitute the points (3, -6) in equation (1), we get 32 + (- 6)2 + 2a(3) + 2b(- 6) + c = 0 9 + 36 + 6a – 12b + c = 0 6a – 12b + c + 45 = 0….. (4) Upon simplifying equations (2), (3), (4) we get a = – 3, b = 1, c = -15. Now by substituting the values of a, b, c in equation (1), we get x2 + y2 + 2(- 3)x + 2(1)y – 15 = 0 x2 + y2 – 6x + 2y – 15 = 0 ∴ The equation of the circle is x2 + y2 – 6x + 2y – 15 = 0. (ii) 2x + y – 3 = 0, x + y – 1 = 0 and 3x + 2y – 5 = 0 Given: The lines 2x + y – 3 = 0 x + y – 1 = 0 3x + 2y – 5 = 0 On solving these lines we get the intersection points A(2, – 1), B(3, – 2), C(1,1) So by using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1) Substitute the points (2, -1) in equation (1), we get 22 + (- 1)2 + 2a(2) + 2b(- 1) + c = 0 4 + 1 + 4a – 2b + c = 0 4a – 2b + c + 5 = 0….. (2) Substitute the points (3, -2) in equation (1), we get 32 + (- 2)2 + 2a(3) + 2b(- 2) + c = 0 9 + 4 + 6a – 4b + c = 0 6a – 4b + c + 13 = 0….. (3) Substitute the points (1, 1) in equation (1), we get 12 + 12 + 2a(1) + 2b(1) + c = 0 1 + 1 + 2a + 2b + c = 0 2a + 2b + c + 2 = 0….. (4) Upon simplifying equations (2), (3), (4) we get a = -13/2, b = -5/2, c = 16 Now by substituting the values of a, b, c in equation (1), we get x2 + y2 + 2 (-13/2)x + 2 (-5/2)y + 16 = 0 ∴ The equation of the circle is x2 + y2 – 13x – 5y + 16 = 0 (iii) x + y = 2, 3x – 4y = 6 and x – y = 0 Given: The lines x + y = 2 3x – 4y = 6 x – y = 0 On solving these lines we get the intersection points A(2,0), B(- 6, – 6), C(1,1) So by using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1) Substitute the points (2, 0) in equation (1), we get 22 + 02 + 2a(2) + 2b(0) + c = 0 4 + 4a + c = 0 4a + c + 4 = 0….. (2) Substitute the point (-6, -6) in equation (1), we get (- 6)2 + (- 6)2 + 2a(- 6) + 2b(- 6) + c = 0 36 + 36 – 12a – 12b + c = 0 12a + 12b – c – 72 = 0….. (3) Substitute the points (1, 1) in equation (1), we get 12 + 12 + 2a(1) + 2b(1) + c = 0 1 + 1 + 2a + 2b + c = 0 2a + 2b + c + 2 = 0….. (4) Upon simplifying equations (2), (3), (4) we get a = 2, b = 3, c = – 12. Substituting the values of a, b, c in equation (1), we get x2 + y2 + 2(2)x + 2(3)y – 12 = 0 x2 + y2 + 4x + 6y – 12 = 0 ∴ The equation of the circle is x2 + y2 + 4x + 6y – 12 = 0 (iv) y = x + 2, 3y = 4x and 2y = 3x Given: The lines y = x + 2 3y = 4x 2y = 3x On solving these lines we get the intersection points A(6,8), B(0,0), C(4,6) So by using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1) Substitute the points (6, 8) in equation (1), we get 62 + 82 + 2a(6) + 2b(8) + c = 0 36 + 64 + 12a + 16b + c = 0 12a + 16b + c + 100 = 0…… (2) Substitute the points (0, 0) in equation (1), we get 02 + 02 + 2a(0) + 2b(0) + c = 0 0 + 0 + 0a + 0b + c = 0 c = 0….. (3) Substitute the points (4, 6) in equation (1), we get 42 + 62 + 2a(4) + 2b(6) + c = 0 16 + 36 + 8a + 12b + c = 0 8a + 12b + c + 52 = 0….. (4) Upon simplifying equations (2), (3), (4) we get a = – 23, b = 11, c = 0 Now by substituting the values of a, b, c in equation (1), we get x2 + y2 + 2(- 23)x + 2(11)y + 0 = 0 x2 + y2 – 46x + 22y = 0 ∴ The equation of the circle is x2 + y2 – 46x + 22y = 0 |
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| 304. |
Find the equation of the circle the end points of whose diameter are the centres of the circles x2 + y2 + 6x – 14y – 1 = 0 and x2 + y2 – 4x + 10y – 2 = 0. |
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Answer» Given: x2 + y2 + 6x – 14y – 1 = 0…. (1) So the centre = [(-6/2), - (-14/2)] = [-3, 7] x2 + y2 – 4x + 10y – 2 = 0… (2) So the centre = [-(-4/2), (-10/2)] = [2, -5] We know that the equation of the circle is given by, (x – x1) (x – x2) + (y – y1) (y – y2) = 0 (x + 3) (x – 2) + (y – 7) (y + 5) = 0 Upon simplification we get x2 + 3x – 2x – 6 + y2 – 7y + 5y – 35 = 0 x2 + y2 + x – 2y – 41 = 0 ∴The equation of the circle is x2 + y2 + x – 2y – 41 = 0 |
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| 305. |
Match the following lists. |
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Answer» Correct Answer - `a rarr r; b rarr p,q; c rarr q,r; d rarr p,s` a. The radical axis of `x^(2)+y^(2)+2a_(1)x+b=0` and `x^(2)+y^(2)+2a_(2)x=b=0` is `(a_(1)-a_(2))x=0` or `x=0` . It must touch both the circles. Solving it with one of the circles, we get `y^(2)+b=0` or `b le 0` b. The radical axis `x^(2)+y^(2)+2a_(1)x+b=0` and `x^(2)+y^(2)+2a_(2)y+b=0` is `a_(1)x-a_(2)y=0` Solving it with one the circles, we have `x^(2)=((a_(1))/(a_(2)))^(2)x^(2)+2a_(1)x+b=0` This equation must have equal roots. Hence, `4a_(1)^(2)-4b[1+((a_(1)^(2))/(a_(2)^(2)))]=0` `a_(1)^(2)-b[1+(a_(1)^(2))/(a_(2)^(2))]=0` Options p and q satisfy this condition. c. If the straight line `a_(1)x-by+b^(2)=0` touches the circle `x^(2)+y^(2)=a_(2)x=by`, then `(|a_(1)(a_(2))/(2)-b(b)/(2)+b^(2)|)/(sqrt(a_(1)^(2)+b^(2)))=sqrt((a_(2)^(2))/(4)+(b^(2))/(4))` `(|a_(1)a_(2)+b^(2)|)/(sqrt(a_(1)^(2)+b^(2)))=sqrt(a_(2)^(2)+b^(2))` ltbgt or `a_(1)^(2)a_(2)^(2)+2b^(2)a_(1)a_(2)+b^(4)=a_(1)^(2)a_(2)^(2)+a_(1)^(2)b^(2)+a_(2)^(2)b^(2)+b^(4)` i.e., `b^(2)=0` or `2a_(1)a_(2) =a_(1)^(2)+a_(2)^(2)` Options q and r satisfy this condition. d. The line `3x+4y-4=0` touches the circle `(x-a_(1))^(2)+(y-a_(2))^(2)=b^(2)`. Then, `(|3a_(1)+4a_(2)-4|)/(5)=b` |
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| 306. |
Match the following lists. |
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Answer» Correct Answer - `a rarr r; b rarr s; c rarr q; d rarr q` a. Since `(2,3)` lies inside the circle, such chord is bisected at (2,3) , which has equation `y-3 = -x(x-2)` or `x+y-5=0` or `a=b=1` b. Let P be the point `(alpha, beta)`. Then `alpha^(2)+beta^(2)+2alpha +2beta =0` The midpoint of OP is `( alpha //2, beta//2)` . Therefore, the locus of `(alpha //2 , beta//2)` is `4x^(2)+4y^(2)+4x+4y=0` i.e., `2g=1,2f=1` and `g+f=1` c. The centes of the circles are (1,2) and (5,-6) The equation of `C_(1)C_(2)` is `y-2= - (8)/(4) (x-1)` i.e., `2x+y-4=0` The equation of the radical axis is `8x-16y-56=0` i.e., `x-2y-7=0` The point of intersection is `(3,-2)`. d. If `theta` is the angle between the tangents, then `(theta)/(2) =("Radius")/("Distance between "(3sqrt(3),3)"and"(0,0))=(1)/(2)` or `(theta)/(2)=(pi)/(6)` or `theta=(pi)/(3)`or `2sqrt(3) tan theta=6` |
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| 307. |
A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation `sqrt3 x+ y -6 = 0` and the point D is (3 sqrt3/2, 3/2). Further, it is given that the origin and the centre of C are on the same side of the line PQ. (1)The equation of circle C is (2)Points E and F are given by (3)Equation of the sides QR, RP areA. `((sqrt(3))/(2),(3)/(2)),(sqrt(3),0)`B. `((sqrt(3))/(2),(1)/(2)),(sqrt(3),0)`C. `((sqrt(3))/(2),(3)/(2)),((sqrt(3))/(2),(1)/(2))`D. `((3)/(2),(sqrt(3))/(2)),((sqrt(3))/(2),(1)/(2))` |
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Answer» Correct Answer - 1 By simple geometry `PD= sqrt(3) ( Delta PQR` is equilateral ) Equation of PQ in parametric form is `(x-(3sqrt(3))/(2))/( cos 120^(@))=(y-(3)/(2))/(sin 120^(@))=r` or `(x-(3sqrt(3))/(2))/( -(1)/(2))=(y-(3)/(2))/((sqrt(3))/(2))=r` Now points at distance `sqrt(3)` from point D on line PQ are given by `((3sqrt(3))/(2) +- sqrt(3) (-(1)/(2)),(3)/(2)+- sqrt(3) ((sqrt(3))/(2)))` or `(2 sqrt(3),0)` and `( sqrt(3),3)` Point C divides the join of P and E in the ratio `2:1` . Similarly, C divides join of Q and F in the ratio `2:1` |
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| 308. |
Consider the family of circles `x^(2)+y^(2)-2x-2ay-8=0` passing through two fixed points A and B . Also, `S=0` is a cricle of this family, the tangent to which at A and B intersect on the line `x+2y+5=0`. If the circle `x^(2)+y^(2)-10x+2y=c=0` is orthogonal to `S=0`, then the value of c isA. 8B. 9C. 10D. 12 |
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Answer» Correct Answer - 4 Circle `x^(2)+y^(2)-10x+2y+c=0` is orthogonal to `x^(2)+y^(2)-2x-6y-8=0`. So, by applying condition of orthogonal intersection, we get `2(-5)(-1)+2(1) (-3)= c-8` `:. C= 12` |
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| 309. |
The sum of the slopes of the lines tangent to both the circles `x^2+y^2=1`and `(x-6)^2+y^2=4`is________ |
| Answer» Since both the circles are symmetric about the y-axis, the sum of slopes of tangest is 0. | |
| 310. |
A circle `x^2+y^2+4x-2sqrt(2)y+c=0`is the director circle of the circle `S_1a n dS_1`is the director circle of circle `S_2,`and so on. If the sum of radii of all these circles is 2, then thevalue of `c`is `ksqrt(2)`, where the value of `k`is___________ |
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Answer» Correct Answer - 4 The radius of the given circle `x^(2)+y^(2)+4x-2sqrt(2)y+c=0` is `ssqrt(4+2-c) = sqrt(6-c)=a` (let). Now, radius of circle `S_(1)=(a)/(sqrt(2))` Radius of circle `S_(2)=(a)/(2)` and so on . Also, `a +(a)/(sqrt(2))+(a)/(2)+ .......+oo=2` (Given ) Then `a((1)/(1-1//sqrt(2)))=2` or `(asqrt(2))/(sqrt(2)-1)=2` or `a=2-sqrt(2)=sqrt(6-c)` or `4+2-4sqrt(2)=6-c` or `c=4 sqrt(2)` |
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| 311. |
consider a family of circles passing through two fixed points `S(3,7)` and `B(6,5)`. If the common chords of the circle `x^(2)+y^(2)-4x-6y-3=0` and the members of the family of circles pass through a fixed point (a,b), then |
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Answer» The equation of the line passing throught the points A(3,7) and B(6,5) is `y-7=-(2)/(3)(x-3)` or `2x+3y-27=0` Also, the equation of the circle with A and B as the endpoints of diameter is `(x-3)(x-6)+(y-7)(y-5)=0` Now, the equation of the family of circles through A and B is `(x-3)(x-6)+(y-7)(y-5)+lambda(2x+3y-27)=0` (1) The equation of the common chrod of (1) and `x^(2)+y^(2)-4x-6y-3=0` is the radical axis, which is `[(x-3)(x-6)+(y-7)(y-5)+lambda(2x+3y-27)]-[x^(2)+y^(2)-4x-6y-3]=0` or `(2 lambda-5)+(3lambda-6)y+(-27lambda+56)=0` or `(-5x-6y+56)+lambda(2x+3y-27)=0` This is the family of lines which passes through the point of intersection of `-5x-6y+56=0` and `2x+3y-27=0`,` i.e., `(2, 23//3)`. |
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| 312. |
The equation of the tangent to the circle `x^2+y^2=25`passing through `(-2,11)`is`4x+3y=25`(b) `3x+4y=38``24 x-7y+125=0`(d) `7x+24 y=250`A. `4x+3y=25`B. `3x+4y=38`C. `24x-7y+125=0`D. `7x+24y=250` |
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Answer» Correct Answer - 1,3 The equation of any tangent to the circle `x^(2)+y^(2)=25` is of form `y= mx + 5 sqrt(1+m^(2))` (where m is the slope ) It passes through `( -2,11) ` .Therefore, `11= - 2m +5 sqrt(1+m^(2))` or `(11+2m)^(2) = 25 (1+m^(2))` or `m=(24)/(7),- (4)/(3)` Thereforem the equation of tangent is `24x-7y+125=0` or `4x+3y=25` |
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| 313. |
Prove that quadrilateral `A B C D`, where `A B-=x+y-10 ,B C-=x-7y+50=0,C D-=22 x-4y+125=0,a n dD A-=2x-4y-5=0,`is concyclic. Also find the equation of the circumcircle of `A B C Ddot` |
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Answer» If ABCD is a cyclic quadrilateral, then the equation of circumcircle of ABCD will be `(x+y-10)(22x-4y+125)+lambda(x-7y+50)(2x-4y-5)=0` Since it represents a circle, we have Coefficient of `x^(2)=` Coefficient of `y^(2)` `implies 22+2 lambda= -4+28 lambda` `implies lambda =1` Also, coefficient of xy is zero `:. 22-4+lambda(-4-14)=0` `:. lambda=1` Thus, value of `lambda` from two conditions is same. Therefore, ABCD is cyclic quadrilateral. Also, equation of circle is `(x+y-10)(22x-4y+125)+(x-7y+50)(2x-4y-5)=0` `:. 24x^(2)+24y^(2)-1500=0` or `2x^(2)+2y^(2)=125` |
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| 314. |
Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic. |
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Answer» Let the equation of the circle passing through the points (3, -2), (1, 0) and (-1, -2) be x2 + y2 + 2gx + 2fy + c = 0 …..(i) For point (3, -2), Substituting x = 3 and y = -2 in (i), we get 9 + 4 + 6g – 4f + c = 0 ⇒ 6g – 4f + c = -13 ….(ii) For point (1, 0), Substituting x = 1 and y = 0 in (i), we get 1 + 0 + 2g + 0 + c = 0 ⇒ 2g + c = -1 ……(iii) For point (-1, -2), Substituting x = -1 and y = -2, we get 1 + 4 – 2g – 4f + c = 0 ⇒ 2g + 4f – c = 5 …….(iv) Adding (ii) and (iv), we get 8g = -8 ⇒ g = -1 Substituting g = -1 in (iii), we get -2 + c = -1 ⇒ c = 1 Substituting g = -1 and c = 1 in (iv), we get -2 + 4f – 1 = 5 ⇒ 4f = 8 ⇒ f = 2 Substituting g = -1, f = 2 and c = 1 in (i), we get x2 + y2 – 2x + 4y + 1 = 0 ……….(v) If (1, -4) satisfies equation (v), the four points are concyclic. Substituting x = 1, y = -4 in L.H.S of (v), we get L.H.S. = (1)2 + (-4)2 – 2(1) + 4(-4) + 1 = 1 + 16 – 2 – 16 + 1 = 0 = R.H.S. Point (1, -4) satisfies equation (v). ∴ The given points are concyclic. |
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| 315. |
In the adjoining figure, `O` is the centre of a circle in which AB and CD are two diameters. Prove that `AC| |BD` and `AD| |BC`. If `angleOBD =50^@`, then find the value of `angleAOC`. |
| Answer» Correct Answer - `80^@` | |
| 316. |
If the lines 2x-3y=5 and 3x-4y=7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.A. `x^(2)+y^(2)+2x-2y=62`B. `x^(2)+y^(2)+2x-2y=47`C. `x^(2)+y^(2)-2x+2y=47`D. `x^(2)+y^(2)-2x+2y=62` |
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Answer» Correct Answer - C Since, 2x-3y = 5 and 3x-4y = 7 are diameters of a circle. Their point of intersection is centre (1, -1) Are given, `pir^(2) = 154` `rArr r^(2) = 154 xx (7)/(22) rArr r = 7 ` ` therefore` Required of circle is `(x-1)^(2) + (y+1)^(2) = 7^(2)` `rArr x^(2) + y^(2)-2x +2y = 47` |
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| 317. |
If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then find the equation of the circle. (A) x2 + y2 – 2x + 2y = 40(B) x2 + y2 – 2x – 2y = 47 (C) x2 + y2 – 2x + 2y = 47 (D) x2 + y2 – 2x – 2y = 40 |
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Answer» Correct option is (C) x2 + y2 – 2x + 2y = 47 Centre of circle = Point of intersection of diameters. Solving 2x – 3y = 5 and 3x – 4y = 7, we get x = 1, y = -1 Centre of the circle C(h, k) = C(1, -1) ∴ Area = 154 πr2 = 154 22/7 x r2 = 154 r2 = 154 x 22/7 = 49 ∴ r = 7 equation of the circle is (x – 1)2 + (y + 1)2 = 72 x2 + y2 – 2x + 2y = 47 |
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| 318. |
The equation(s) of the tangent(s) to the circle x2 + y2 = 4 which are parallel to x + 2y + 3 = 0 are (A) x – 2y = 2 (B) x + 2y = ±2√3 (C) x + 2y = ±2√5 (D) x – 2y = ±2√5 |
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Answer» Correct option is : (C) x + 2y = ±2√5 |
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| 319. |
Find the equation of the circle which passes through the points `(1,-2),(4,-3)`and whose center lies on the line `3x+4y=7.` |
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Answer» Let the equation of the circle be `x^(2)+y^(2)+2gx+2fy+c=0` (1) If (1) passes through the points (1,-2) and (4,-3), then `5+2g-4f+c=0` (2) and `25+8g-6f+c=0 ` (3) Since the center `(-g,-f)` lies on the line `3x+4y=7`, we have `-3g-4f=7` (4) Solving (2), (3), and (4), we get `g=-(47)/(15),f=(3)/(5)`, and `c=(11)/(3)` Substituting in (1), the equation of the circle is `15x^(2)+15y^(2)-94x+18y+55=0` |
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| 320. |
Fill in the blanks :(i) Center of circle lies in ………… of circle. (exterior/interior)(ii) A point, having distance from center greater than radius lies in the ……… of the circle (exterior/interior)(iii) Longest chord of circle is ………… of the circle.(iv) An arc is …………. when its ends are ends of diameter.(v) A circle lies on which plane divides it into ………… parts. |
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Answer» (i) Center of circle lies in interior of circle. (ii) A point, having distance from center greater then radius lies in the exterior of the circle. (iii) Longest chord of circle is diameter of the circle. (iv) An arc is semicircle when its ends are ends of diameter. (v) A circle lies on which plane divides it into three parts. |
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| 321. |
Find the equation of the circle which passes through the points (2, 3) and (4, 5), and the center lies on the straight line y – 4x + 3 = 0. (A) x2 + y2 - 4x - 10y + 25 = 0 (B) x2 + y2 -4x - 10y – 25 = 0 (C) x2 + y2-4x + 10y – 25 = 0 (D) x2 + y2 + 4x - 10y + 25 = 0 |
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Answer» Correct option is : (A) x2 + y2 - 4x - 10y + 25 = 0 |
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| 322. |
If the lines `x+y=6a n dx+2y=4`are diameters of the circle which passes through the point (2, 6), then findits equation. |
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Answer» Here center will be the point of intersection of the diameter, i.e., `C (8,-2)`. Also, the circle passes through the point P(2,6). Then radius is `CP=10`. Hence, the required equation is `(x-8)^(2)+(y+2)^(2)=10^(2)` or `x^(2)+y^(2)-16x+4y-32=0` |
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| 323. |
Find the equation of the circle with radius 5 whose center lies on thex-axis and passes through the point (2, 3). |
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Answer» Since the radius of the circle is 5 and its center lies on the x-axis, the equation of the circle is `(x-h)^(2)+y^(2)=25`. It is given that the circle passes through the point (2,3). Therefore, `(2-h)^(2)+3^(2)=25` or `(2-h^(2))=16` or `2-h=+-4` If `2-h=4,` then `h=2` If `2-h= -4` then `h=6` Therefore , the equation of circle is `(x+2)^(2)+y^(2)=25` or `(x+6)^(2)+y^(2)=25`. Hence, `x^(2)+y^(2)+4x-21=0` or `x^(2)+y^(2)-12x+11=0` |
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| 324. |
In the given figure, O is the centre of the circle whose diameter is MN. Measures of some central angles are given in the figure. i. m∠AOB and m∠COD ii. Show that arc AB ≅ arc CD iii. Show that chord AB ≅ chord CD |
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Answer» i. Seg MN is the diameter of the circle. … [Given] ∴ m∠AOM + m∠AON = 180° … [Angles in a linear pair] ∴ m∠AOM + (m∠AOB + m∠BON) = 180° … [Angle addition property] ∴ 100° + m∠AOB + 35° = 180° …[∵ m∠AOM = 100°, m∠BON = 35°] ∴ m∠AOB + 135° = 180° ∴ m∠AOB = 180°- 135° ∴m∠AOB = 45° …(i) Also, m∠DOM + m∠DON = 180° … [Angles in a linear pair] ∴ m∠DOM + (m∠COD + m∠CON) = 180° … [Angle addition property] ∴ 100° +m∠COD + 35°= 180° …[∵ m∠DOM = 100°, m∠CON = 35° ] ∴ m∠COD + 135° = 180° ∴ m∠COD = 180°- 135° ∴ m∠COD = 45° …(ii) ii. m(arc AB) = m∠AOB = 45° … [From (i)] m(arc DC) = m∠DOC = 45° .. .[From (ii)] ∴ m(arc AB) = m(arc DC) …[From (i) and (ii)] ∴ arc AB ≅ arc CD … [If the measures of two arcs of a circle are same, then the two arcs are congruent] iii. arc AB ≅ arc CD ∴ chord AB ≅ chord CD ….[The chords corresponding to congruent arcs are congruent] |
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| 325. |
MN is tangent at M and AM is radius . Find AM. A. 6B. 3C. `3sqrt(3)`D. 1 |
| Answer» Correct Answer - C | |
| 326. |
The least distance of the line `8x -4y+73=0` from the circle `16x^2+16y^2+48x-8y-43=0` is |
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Answer» Correct Answer - `2sqrt(5)` `16x^(2)+16y^(2)+48x-8y-43=0` or `x^(2)+y^(2)+3x-(1)/(2)y-(43)/(16)=0` Center of the circle `: (-(3)/(2),(1)/(4))` Radius`= sqrt((9)/(4)+(1)/(16)+(43)/(16))=sqrt(5)` Required least distance `=` Distance of line from centre of circle `-` Radius of circle `= (|8(-(3)/(2))-4((1)/(4))+73|)/(sqrt(64+16))-sqrt(5)` `=(60)/(4 sqrt(5))-sqrt(5)= 2 sqrt(5)` |
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| 327. |
The length of the tangent from any point on the circle to the circle `(x-3)^2 + (y + 2)^2 =5r^2` to the circle ` (x-3)^2 + (y + 2)^2 = r^2` is 4 units. Then the area between the circles is |
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Answer» Correct Answer - `16 pi` From the figure, `5r^(2)=16+r^(2)` `:. R^(2)= 4` `:. R = 2` `:. ` Area between circles `=pi(5r^(2))=pi(r^(2))=4pir^(2)=16pi` |
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| 328. |
In the given figure, there are two circles with centers A and B. The common tangent to the circles touches them, respectively,at P and Q. AR is 40cm and AB is divided by the point of contact of the circles in the ratio `5:3` The radius of the circle with center B isA. 10 cmB. 3 cmC. 6cmD. 8cm |
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Answer» Correct Answer - 3 `(r_(1))/(r_(2))=(5)/(3)` `:. (r_(1)+r_(2))/(r_(2))=(8)/(3)` `:. r_(2)=((40-24)3)/(8)=6 cm` |
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| 329. |
In the given figure, there are two circles with centers A and B. The common tangent to the circles touches them, respectively,at P and Q. AR is 40cm and AB is divided by the point of contact of the circles in the ratio `5:3` What is the ratio of the length of AB to that of BR ?A. `1:4`B. `2:3`C. `2:5`D. `7:4` |
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Answer» Correct Answer - 2 `(r_(1))/(r_(2))=(5)/(3)=(AR)/(BR)` or `BR=(3 xx 40)/(5) =25 ` cm `(AB)/(BR)=(40-24)/(24)=(16)/(24)=(2)/(3)` |
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| 330. |
If the length tangent drawn fromthe point (5, 3) to the circle `x^2+y^2+2x+k y+17=0`is 7, then find the value of `kdot` |
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Answer» Correct Answer - `-4` According to the question, `sqrt((5)^(2)+(3)^(2)+2(5)+k(3)+17)=7` or `61+3k=49` or `k= -4` |
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| 331. |
In the given figure, there are two circles with centers A and B. The common tangent to the circles touches them, respectively,at P and Q. AR is 40cm and AB is divided by the point of contact of the circles in the ratio `5:3` The length of QR isA. `10sqrt(15)` cmB. `5 sqrt(15) cm`C. `4 sqrt(15)` cmD. `6 sqrt(15)` cm |
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Answer» Correct Answer - 4 `QR =sqrt((BR)^(2)-r_(2)^(2))=sqrt((24)^(2)-36)=6 sqrt(15) cm` |
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| 332. |
Let each of the circles `S_(1)-=x^(2)+y^(2)+4y-1=0` `S_(1)-= x^(2)+y^(2)+6x+y+8=0` `S_(3)-=x^(2)+y^(2)-4x-4y-37=0` touch the other two. Also, let `P_(1),P_(2)` and `P_(3)` be the points of contact of `S_(1)` and `S_(2) , S_(2)` and `S_(3)`, and `S_(3)` , respectively, `C_(1),C_(2)` and `C_(3)` are the centres of `S_(1),S_(2)` and `S_(3)` respectively. The ratio `("area"(DeltaP_(1)P_(2)P_(3)))/("area"(DeltaC_(1)C_(2)C_(3)))` is equal toA. `3:2`B. `2:3`C. `5:3`D. `2:5` |
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Answer» Correct Answer - 4 `S_(1)-= x^(2)+y^(2)+4y-1=0` `S_(2) -= x^(2)+y^(2)+6x+y+8=0` `S_(3)-=x^(2)+y^(2)-4x-4y-37=0` `C_(1) -= (0,-2),r_(1)=sqrt(5)` `C_(2) -= (-3,(-1)/(2)),r_(2)=(sqrt(5))/(2)` `C_(3)-= (2,2), r_(3)= 3 sqrt(5)` Also, `C_(1)C_(2)=sqrt(9+(9)/(4))=(3sqrt(5))/(2)=r_(1)+r_(2)` So, `S_(1)` and `S_(2)` touch each other externally, `C_(2)C_(3)=sqrt(25+(25)/(4))=(5sqrt(5))/(2)=r_(3)-r_(2)` So, `S_(2)` and `S_(3)` touch each other internally. The point of contact `P_(1)` divides `C_(1)C_(2)` internally in the ratio `r_(1) : r_(2) = 2:1` `implies P_(1) -= (-2 ,-1)` The point of contact `P_(2)` divides `C_(2)C_(3)` externally in the ratio `r_(2) : r_(3) = 1:6` `implies P_(2) -= (-4, -1)`. The point of contact `P_(3)` divides `C_(3)C_(1)` externally in the ratio `r_(3) : r_(1) = 3:1` `implies P_(3) -= ( -1,-4)` Area of `Delta P_(1)P_(2)P_(3)= (1)/(2) | {:(-2,-1,1),(-4,-1,1),(-1,-4,1):}| = 3` And area `Delta C_(1)C_(2)C_(3)= (1)/(2) |{:(0,-2,1),(-3,(-1)/(2),1),(2,2,1):}|=(15)/(2)` `:. ("area "(Delta P_(1)P_(2)P_(3)))/("area"(DeltaC_(1)C_(2)C_(3)))=(3)/(15)=2:5` Clearly, `P_(2)(-4,-1)` and `P_(3)(-1,-4)` are images of each other with respect to the line `y=x` |
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| 333. |
Let each of the circles `S_(1)-=x^(2)+y^(2)+4y-1=0` `S_(1)-= x^(2)+y^(2)+6x+y+8=0` `S_(3)-=x^(2)+y^(2)-4x-4y-37=0` touch the other two. Also, let `P_(1),P_(2)` and `P_(3)` be the points of contact of `S_(1)` and `S_(2) , S_(2)` and `S_(3)`, and `S_(3)` , respectively, `C_(1),C_(2)` and `C_(3)` are the centres of `S_(1),S_(2)` and `S_(3)` respectively. The coordinates of `P_(1)` areA. `(2,-1)`B. `(-2,-1)`C. `(-2,1)`D. `(2,1)` |
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Answer» Correct Answer - 2 `S_(1)-= x^(2)+y^(2)+4y-1=0` `S_(2) -= x^(2)+y^(2)+6x+y+8=0` `S_(3)-=x^(2)+y^(2)-4x-4y-37=0` `C_(1) -= (0,-2),r_(1)=sqrt(5)` `C_(2) -= (-3,(-1)/(2)),r_(2)=(sqrt(5))/(2)` `C_(3)-= (2,2), r_(3)= 3 sqrt(5)` Also, `C_(1)C_(2)=sqrt(9+(9)/(4))=(3sqrt(5))/(2)=r_(1)+r_(2)` So, `S_(1)` and `S_(2)` touch each other externally, `C_(2)C_(3)=sqrt(25+(25)/(4))=(5sqrt(5))/(2)=r_(3)-r_(2)` So, `S_(2)` and `S_(3)` touch each other internally. The point of contact `P_(1)` divides `C_(1)C_(2)` internally in the ratio `r_(1) : r_(2) = 2:1` `implies P_(1) -= (-2 ,-1)` The point of contact `P_(2)` divides `C_(2)C_(3)` externally in the ratio `r_(2) : r_(3) = 1:6` `implies P_(2) -= (-4, -1)`. The point of contact `P_(3)` divides `C_(3)C_(1)` externally in the ratio `r_(3) : r_(1) = 3:1` `implies P_(3) -= ( -1,-4)` Area of `Delta P_(1)P_(2)P_(3)= (1)/(2) | {:(-2,-1,1),(-4,-1,1),(-1,-4,1):}| = 3` And area `Delta C_(1)C_(2)C_(3)= (1)/(2) |{:(0,-2,1),(-3,(-1)/(2),1),(2,2,1):}|=(15)/(2)` `:. ("area "(Delta P_(1)P_(2)P_(3)))/("area"(DeltaC_(1)C_(2)C_(3)))=(3)/(15)=2:5` Clearly, `P_(2)(-4,-1)` and `P_(3)(-1,-4)` are images of each other with respect to the line `y=x` |
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| 334. |
Let each of the circles `S_(1)-=x^(2)+y^(2)+4y-1=0` `S_(1)-= x^(2)+y^(2)+6x+y+8=0` `S_(3)-=x^(2)+y^(2)-4x-4y-37=0` touch the other two. Also, let `P_(1),P_(2)` and `P_(3)` be the points of contact of `S_(1)` and `S_(2) , S_(2)` and `S_(3)`, and `S_(3)` , respectively, `C_(1),C_(2)` and `C_(3)` are the centres of `S_(1),S_(2)` and `S_(3)` respectively. `P_(2)` and `P_(3)` are images of each other with respect to the lineA. `y=x`B. `y = -x`C. `y=x+1`D. `y= -x+2` |
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Answer» Correct Answer - 1 `S_(1)-= x^(2)+y^(2)+4y-1=0` `S_(2) -= x^(2)+y^(2)+6x+y+8=0` `S_(3)-=x^(2)+y^(2)-4x-4y-37=0` `C_(1) -= (0,-2),r_(1)=sqrt(5)` `C_(2) -= (-3,(-1)/(2)),r_(2)=(sqrt(5))/(2)` `C_(3)-= (2,2), r_(3)= 3 sqrt(5)` Also, `C_(1)C_(2)=sqrt(9+(9)/(4))=(3sqrt(5))/(2)=r_(1)+r_(2)` So, `S_(1)` and `S_(2)` touch each other externally, `C_(2)C_(3)=sqrt(25+(25)/(4))=(5sqrt(5))/(2)=r_(3)-r_(2)` So, `S_(2)` and `S_(3)` touch each other internally. The point of contact `P_(1)` divides `C_(1)C_(2)` internally in the ratio `r_(1) : r_(2) = 2:1` `implies P_(1) -= (-2 ,-1)` The point of contact `P_(2)` divides `C_(2)C_(3)` externally in the ratio `r_(2) : r_(3) = 1:6` `implies P_(2) -= (-4, -1)`. The point of contact `P_(3)` divides `C_(3)C_(1)` externally in the ratio `r_(3) : r_(1) = 3:1` `implies P_(3) -= ( -1,-4)` Area of `Delta P_(1)P_(2)P_(3)= (1)/(2) | {:(-2,-1,1),(-4,-1,1),(-1,-4,1):}| = 3` And area `Delta C_(1)C_(2)C_(3)= (1)/(2) |{:(0,-2,1),(-3,(-1)/(2),1),(2,2,1):}|=(15)/(2)` `:. ("area "(Delta P_(1)P_(2)P_(3)))/("area"(DeltaC_(1)C_(2)C_(3)))=(3)/(15)=2:5` Clearly, `P_(2)(-4,-1)` and `P_(3)(-1,-4)` are images of each other with respect to the line `y=x` |
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| 335. |
Prove that the equation `x^(2)+y^(2)-2x-2ay-8=0, a in R ` represents the family of circles passing through two fixed points on x-axis. |
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Answer» We have `x^(2)+y^(2)-2x-2ay-8=0`. `:. (x^(2)+y^(2)-2x-8)-2a(y)=0,a in R` This equation is of the form `S+ lambda L=0`, which represents family of circle passing through points of intersection of circle `x^(2)+y^(2)-2x-8=0` and line `y=0` (x-axis). Solving circle and line, we have `x^(2)-2x-8=0` `implies (x-4)(x+2)=0` `implies x=4,x= -2` So, circle pass through two fixed points (4,0) and (-2,0) on x-axis. |
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| 336. |
Consider the family of circles `x^(2)+y^(2)-2x-2ay-8=0` passing through two fixed points A and B . Also, `S=0` is a cricle of this family, the tangent to which at A and B intersect on the line `x+2y+5=0`. The distance between the points A and B , isA. 4B. `4 sqrt(2)`C. 6D. 8 |
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Answer» Correct Answer - 3 We have `x^(2)+y^(2) -2x-2ay-8=0` or `(x^(2)+y^(2)-2x-8)-2ay =0`, which is of the form `S + lambda L =0`. Solving `S -= x^(2)+y^(2) -2x-8=0` and `L -= y =0`, we get `x^(2)-2x-8=0` `implies (x-4) (x+2)=0` `implies x = - 2` or `4` `implies A -= (4,0) , B -= (-2,0)` Distance between A and `B =6` |
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| 337. |
Find the equation of the circle which passes through the origin and cuts off intercepts a and b respectively from x and y – axes. |
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Answer» Since the circle has intercept ‘b’ from x – axis, the circle must pass through (0, b) and (0, -b) as it already passes through the origin. Let us assume the circle passing through the points A(a,0) and B(0,b). We know that the equation of the circle with AB as diameter is given by (x – x1) (x – x2) + (y – y1) (y – y2) = 0 (x – a) (x – 0) + (y – 0) (y – b) = 0 x2 + y2 + ax + by = 0 or x2 + y2 – ax – by = 0 ∴The equation of the circle is x2 + y2 + ax + by = 0 or x2 + y2 – ax – by = 0 |
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| 338. |
The centre of the smallest circle touching the circles `x^2+ y^2-2y -3=0` and `x^2 +y^ 2-8x -18y +93= 0` is:A. (3,2)B. (4,4)C. (2,5)D. (2,7) |
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Answer» Correct Answer - 3 Circles `S_(1) : x^(2)+y^(2)-2y-3=0` and `S_(2) : x^(2)+y^(2)-8x-18y+93=0` `C_(1) -= (0,1)` and `r_(1)=2` `C_(2) -= (4,9)` and `r_(2)=2` Since circles have same radius , centre `C_(3)` of the smallest circle is collinear with `C_(1)` and `C_(2)` and is midpoint of `C_(1)C_(2)`. `:. C_(3) -= (2,5)` |
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| 339. |
The circles `x^2+y^2-2x-4y+1=0`and `x^2+y^2+4x+4y-1=0`touch internallytouch externallyhave `3x+4y-1=0`as the common tangent at thepoint of contacthave `3x+4y+1=0`as the common tangent at thepoint of contactA. touch internallyB. touch externallyC. have `3x+4y-1=0` as the common tangent at the point of contactD. have `3x+4y+1=0` as the common tangent at the point of contanct. |
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Answer» Correct Answer - 2,3 For the given circles `S_(1) : x^(2)+y^(2)-2x-4y+1=0` and `S_(2) : x^(2) + y^(2)+4x+4y-1=0` `C_(1) -= (1,2) r_(1)=2,C_(2) -= (-2,-2) , ` and `r_(2)=3` Now, `r_(1)+r_(2)=5` and `C_(1)C_(2)=5`. Hence, the circles touch externally. Also, the common tangent at the point of contact is `S_(1) - S_(2)=0` or `3x+4y-1=0`. |
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| 340. |
Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent : x2 + y2 – 4x – 4y – 28 = 0, x2 + y2 – 4x – 12 = 0 |
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Answer» Given equation of the first circle is x2 + y2 – 4x – 4y – 28 = 0 Here, g = -2, f = -2, c = -28 Centre of the first circle is C = (2, 2) Radius of the first circle is r1 = \(\sqrt{(-2)^2 + (-2)^2+28}\) \(\sqrt{4+ 4+28}\) = √36 = 6 Given equation of the second circle is x2 + y2 – 4x – 12 = 0 Here, g = -2, f = 0, c = -12 Centre of the second circle is C2 = (2, 0) Radius of the second circle is r2 = \(\sqrt {(-2)^2 + 0^2 + 12}\) = \(\sqrt {4+ 12}\) = 4 By distance formula, C1 C1 = \(\sqrt{(2-2)^2 + (0-2)^2}\) = √4 = 2 r1 – r2 | = 6 – 4 = 2 Since, C1 C2 = |r1 – r2 | ∴ the given circles touch each other internally. Equation of common tangent is (x2 + y2 – 4x – 4y – 28) – (x2 + y2 – 4x – 12) = 0 ⇒ -4x – 4y – 28 + 4x + 12 = 0 ⇒ -4y – 16 = 0 ⇒ y + 4 = 0 ⇒ y = -4 Substituting y = -4 in x2 + y2 – 4x – 12 = 0, we get ⇒ x2 + (-4)2 – 4x – 12 = 0 ⇒ x2 + 16 – 4x – 12 = 0 ⇒ x2 – 4x + 4 = 0 . ⇒ (x – 2)2 = 0 ⇒ x = 2 ∴ Point of contact is (2, -4) and equation of common tangent is y + 4 = 0. |
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| 341. |
The equation of the line passing through the points of intersection of the circles `3x^2 +3y^2-2x + 12y-9=0 and x^2+y^2+6x+2y-15=0` is |
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Answer» Equation of straight line passing through intersection of two circles `C_(1) and C_(2)` is `(S_(1)-S_(2))=0` `therefore x^(2)+y^(2)-(2)/(3)x+4y-3-(x^(2)+y^(2)+6x+2y-15)=0` `rArr (20)/(3)x-2y-12=0 rArr10x-3y-18=0` |
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| 342. |
The equation of circle concentric with circle `x^2+y^2-6x+ 12y+ 15=0` and double its area is |
| Answer» Correct Answer - ` x ^(2) + y ^(2) - 6x + 12y - 15 = 0` | |
| 343. |
Obtain the parametric equation of the circle represented by `x^(2) + y^(2) + 6x + 8y- 96= 0`A.B.C.D. |
| Answer» Correct Answer - `0 le theta le 2pi` | |
| 344. |
Let RS be the diameter of the circle `x^2+y^2=1,` where S is the point `(1,0)` Let P be a variable apoint (other than `R and S`) on the circle and tangents to the circle at `S and P` meet at the point Q.The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. then the locus of E passes through the point(s)- (A) `(1/3,1/sqrt3)` (B) `(1/4,1/2)` (C) `(1/3,-1/sqrt3)` (D) `(1/4,-1/2)`A. `((1)/(3),(1)/(sqrt3))`B. `((1)/(4),(1)/(2))`C. `((1)/(3),-(1)/(sqrt3))`D. `((1)/(4),-(1)/(2))` |
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Answer» Correct Answer - A::C |
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| 345. |
Length of a tangent segment drawn from a point which is at a distnace 12.5 cm from the centre of a circle is 12cm , find the diameter of the circle.A. 25 cmB. 24 cmC. 7 cmD. 14 cm |
| Answer» Correct Answer - C | |
| 346. |
AP is tangent to the circle with centre O at point A. OP =10 cm and `/_ OPA = 30^(@)`. The radiu of the circle isA. 10 cmB. `5 sqrt(3) `cmC. 5 cmD. `10 sqrt(3) cm ` |
| Answer» Correct Answer - C | |
| 347. |
In the figure, chords AD and BC intersect each other at right angles at a point P. If `/_DAB = 35^(@)`, then `/_ADC =` A. `35^(@)`B. `55^(@)`C. `65^(@)`D. `45^(@)` |
| Answer» Correct Answer - B | |
| 348. |
In the figure, if `/_ABC = 60^(@)` , then `/_AOC = ` A. `120^(@)`B. `60^(@)`C. `30^(@)`D. `90^(@)` |
| Answer» Correct Answer - A | |
| 349. |
In the adjoining figure, `O` is the centre of the circle. If `angleAOB=70^@`, then find the value of `angle OCD`. |
| Answer» Correct Answer - `55^@` | |
| 350. |
Name the shape shown in centre of our national flag. In how many parts it is divided ? Also explain the value shown. |
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Answer» Correct Answer - It is a circle (Ashok Chakra) divided by 24 spoke wheel. Value: it gives us inspiration of pathriotic and keep telling that the should be alert 24 hours for ourself, our society and for our nation. |
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