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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.
| 151. |
In the given figure, PA and PB are tangents to the given circle such that PA = 5 cm and ∠APB = 60° . The length of chord AB is(a) 5√2 cm(b) 5 cm(c) 5√3 cm(d) 7.5 cm |
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Answer» Correct answer is (b) 5 cm The lengths of tangents drawn from a point to a circle are equal So, PA = PB and therefore, ∠PAB = ∠PBA = x (say). Then, in △PAB : ∠PAB + ∠PBA + ∠APB = 180° \(\Rightarrow\) x + x + 60° = 180° \(\Rightarrow\) 2x = 180° - 60° \(\Rightarrow\) 2x = 120° \(\Rightarrow\) x = 60° ∴ Each angle of △PAB is 60° and therefore, it is an equilateral triangle. ∴ AB = PA = PB = 5 cm ∴ The length of the chord AB is 5 cm. |
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| 152. |
In Fig. 9.2, PQ is a chord of a circle and PT is the tangent at P such that ∠QPT = 60°. Then ∠PRQ is equal to(A) 135° (B) 150° (C) 120° (D) 110° |
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Answer» correct answer is (C) 120° [Hint : ∠OPQ = ∠OQP = 30°, i.e., ∠POQ = 120°. Also, ∠PRQ = 1/2 reflex ∠POQ] |
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| 153. |
If angle between two tangents drawn from a point P to a circle of radius a and centre O is 60°, then OP = a√3 . |
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Answer» If angle between two tangents drawn from a point P to a circle of radius a and centre O is 60°, then OP = a√3 . False |
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| 154. |
Thenumber of common tangents to the circles `"x"^("2")"+y"^("2")"-4x-6y-12=0"`and `x^2+""y^2+""6x""+""18 y""+""26""=""0`,is :(1)1 (2) 2 (3) 3 (4) 4 |
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Answer» `x^2 + y^2 - 4x - 6y - 12= 0` `x^2 - 4x + 4 + y^2 - 6y+ 9 - 12 - 4 - 9 = 0` `(x-2)^2 + (y-3)^2 = 25= 5^2` `c_1 = (2,3) & r_1= 5` `x^2 + y^2 + 6x + 18y + 26= 0` `x^2 + 6x + 9 + y^2 + 18y + 81 + 26 - 9 - 81= 0` `(x+3)^2 + (y+9)^2 = 64= 8^2` `c_2 = (-3,-9) & r_2= 8` `c_1 = (2,3); r_1=5` `c_2 = (-3,-9) ; r_2= 8` `d(c_1,c_2) = sqrt((2- (-3))^2 + (3- (-9))^2` `= sqrt(25 + 144) = sqrt(169)= 13` `r_1 = 5 ; r_2= 8` `r_1 + r_2 = 13` so,`3 `tangents option 2 is correct |
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| 155. |
There are two circles whose equation are `x^2+y^2=9`and `x^2+y^2-8x-6y+n^2=0,n in Zdot`If the two circles have exactly two common tangents, then the number ofpossible values of `n`is2 (b) 8(c) 9 (d)none of theseA. 2B. 8C. 9D. none of these |
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Answer» Correct Answer - C The coordinates of the centres and radii of the circles are: `{:("Centres:",C_(1)(0, 0),C_(2)(4, 3)),("Radii",r_(1)=3,r_(2)=sqrt(25-n^(2))-5 lt n lt 5),(,,):}` Given circles will have exactly two common tangents, if `|r_(1)-r_(2)| lt C_(1) C_(2) lt R_(1)+r_(2)` `rArr |3-sqrt(25-n^(2))| lt 5` is true for all `n in (-5, 5)`. Now, `5 lt 3 + sqrt(25-n^(2)) ` `rArr 2 lt sqrt(25-n^(2))` `rArr 4 lt 25 - n^(2)` `rArr n^(2)-21 lt 0 ` `rArr -sqrt (21) lt n lt sqrt(21) rArr n = pm 4, pm 3, pm 2, pm1, 0`. Hence, n can take 9 integral values. |
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| 156. |
If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents at A and B is also 60°. |
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Answer» If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents at A and B is also 60°. False |
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| 157. |
Find the number of commontangents that can be drawn to the circles `x^2+y^2-4x-6y-3=0`and `x^2+y^2+2x+2y+1=0`A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - C | |
| 158. |
Find the equation of the circle whose radius is 3 and which touchesinternally the circle `x^2+y^2-4x-6y=-12=0`at the point `(-1,-1)dot`A. `5x^(2)+5y^(2)+8x -14y-16=0`B. `5x^(2)+5y^(2)-8x-14y-32=0`C. `5x^(2)+5y^(2)-8x+14y-4=0`D. `5x^(2)+5y6(2)+8x+14y+12=0` |
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Answer» Correct Answer - B The given circle `x^(2)+y^(2)-4x-6y-12=0` has its centre at (2, k) and radius equal to 5. Let (h, k) be the centre of the required circle. Then, the point (h, k) divides the line joining (-1, -1) to (2, 3) in the ratio 3:2, where 3 is the radius of the required circle. Thus, we have `h=(3xx2+2xx(-1))/(3+2)=(4)/(5) and k=(3xx3+2xx(-1))/(3+2)=(7)/(5)` Hence, the equation of the reuired circle is `(x-(4)/(5))^(2)+(y-(7)/(5))^(2)=3^(2)` `rArr 5x^(2)+5y^(2)-8x-14y-32=0` |
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| 159. |
In two concentric circles prove that all chords of the outer circle which touch the inner circle are of equal length. |
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Answer» `In/_AOC` `OC^2+AC^2=OA^2` `r_1^2+AC^2=r_2^2` `AC=sqrt(r_2^2-r_1^2)` `AB=2sqrt(r_2^2-r_1^2)` AC=BC AB`_|_`BC C is mid point of AB length of chord is constant =`2sqrt(r_2^2-r_1^2)`. |
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| 160. |
Consider the circles `x^2+(y-1)^2=9,(x-1)^2+y^2=25.`They are such thatthese circles touch each otherone of these circles liesentirely inside the othereach of these circles lies outside the otherthey intersect at two points.A. these circles touch each otherB. one of these circles lies entirely inside the otherC. each of these circles lies outside the otherD. they intersect in two point |
| Answer» Correct Answer - B | |
| 161. |
Find the angle which the common chord of `x^2+y^2-4x=0`and `x^2+y^2=16`subtends at the origin.A. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)` |
| Answer» Correct Answer - D | |
| 162. |
The point of which the line `9x + y - 28 = 0` is the chord of contact of the circle `2x^2+2y^2-3x+5y-7=0` isA. (3, 1)B. (1, 3)C. (3, -1)D. (-3, 1) |
| Answer» Correct Answer - C | |
| 163. |
If a chord of contact of tangents drawn from a point P with respect to the circle `x^(2)+y^(2)=9` is x=2, then area, in square units, of triangle formed by tangents drawn from P to the circle and their chord of contact is equal toA. `(4 sqrt(5))/(2)`B. `(9sqrt(3))/(2)`C. `5sqrt(5))/(2)`D. none of these |
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Answer» Correct Answer - C Let `(x_(1),y_(1))` be the coordinates of P. Then, the equation of AB is `x x_(1)+y y_(1)=9`. Clearly, this equation and x=2 represent the same line `:. (x_(1))/(1)=(y_(1))/(0)=(9)/(2)rArr x_(1)=(9)/(2), y_(1)=0`. We know that the area of the triangle formed by the tangents drawn from a point `P(x_(1), y_(1))` to the circle `x^(2)+y^(2)=a^(2)` and their chord of contact is `(a(x_(1)^(2)+y_(1)^(2)-a^(2))^(3//2))/(x_(1)^(2)+y_(1)^(2))` `:. ` Required area `=(3((81)/(4)+0-9)^(3//3))/((81)/(4)+0)` sq. units`=(5sqrt(5))/(2)`sq. units |
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| 164. |
The two circles `x^(2)+y^(2)-2x-3=0` and `x^(2)+y^(2)-4x-6y-8=0` are such thatA. they touch each otherB. they intersect each otherC. one lies inside the otherD. each lies outside the other |
| Answer» Correct Answer - B | |
| 165. |
The equation of the circle on the common chord of the circles `(x-a)^(2)+y^(2)=a^(2)` and `x^(2)+(y+b)^(2)=b^(2)` as diameter, isA. `x^(2)+y^(2)=2ab(bx+ay)`B. `x^(2)+y^(2)=bx+ay)`C. `(a^(2)+b^(2))(x^(2)+y^(2))=2ab(bx-ay)`D. `(a^(2)+b^(2))(x^(2)+y^(2))=2(bx+ay)` |
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Answer» Correct Answer - C The equation of the common chord of the circles `(x-a)^(2)+y^(2)=a^(2) and x^(2)+(y+b)^(2)=b^(2)` is `ax+by=0 ` ...(i) The equation of the required circle is `{(x-a)^(2)+y^(2)-a^(2)} + lambda(ax+by)=0` Since (i) is a diameter of circle (ii). `:. a{-((a lambda-2a))/(2)}+b (-(lambda b)/(2))=0 rArr lambda = (2^(2))/(a^(2)+b^(2))` Putting the value of `lambda` in (ii), the equation of the required circle is `(a^(2)+b^(2))(x^(2)+y^(2))=2ab(bx-ay)` |
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| 166. |
The equation of the circle whose diameter is the common chord of thecircles; `x^2+y^2+3x+2y+1=0`& `x^2+y^2+3x+4y+2=0`is:`x^2+y^2+8x+10 y+2=0``x^2+y^2-5x+4y+7=0``2x^2+2y^2+6x+2y+1=0`None of theseA. `x^(2)+y^(2)+8x+10y+2=0`B. `x^(2)+y^(2)-5x+4y+7=0`C. `2x^(2)+2y^(2)+6x+2y+1=0`D. none of these |
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Answer» Correct Answer - C The equation of the common chord of the circles `x^(2)+y^(2)+3x+2y+1=0 and x^(2)+y^(2)+3x+4y+2=0` is given by 2y+1=0 [Using : `S_(1)-S_(2)=0`] The equation of a circle passing through the intersection of the given circles is `(x^(2)+y^(2)+3x+2y+1)+lambda(x^(2)+y^(2)+3x+4y+2)=0` `rArr x^(2)+y^(2)+x((3+3lambda)/(lambda+1))+2lambda((1+2lambda)/(lambda+1))+(1+2lambda)/(lambda+1)=0` ...(i) Since 2 y+1=0 is a diameter of this circle. Therefore, its centre `(-(3)/(2), -(2lambda+1)/(lambda + 1))` lies on it. `:. - 2 ((2lambda+1)/(lambda+1))+1=0rArr - 4 lambda-2 + lambda+1=0rArr lambda=-(1)/(3)` Putting `lambda=-1//3` in (i), the equation of the required circle is `2x^(2)+2y^(2)+6x+2y+1=0` |
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| 167. |
The length of the common chord of the circles `x^(2)+y^(2)-2x-1=0` and `x^(2)+y^(2)+4y-1=0`, isA. `sqrt(15//2)`B. `sqrt(15)`C. `2sqrt(15)`D. none of these |
| Answer» Correct Answer - A | |
| 168. |
The point of intersection of the common chords of three circles described on the three sides of a triangle as diameter isA. centroid of the triangleB. orthocentre of the triangleC. circumcentre of the triangleD. incentre of the triangle |
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Answer» Correct Answer - B Let ABC be a triangle and let AD, BE and CF be its altitudes. Then, `angleADB=pi//2` and `angleAEB=pi//2`. Therefore, the circle on AB and AC as diameters passes through D. Consequently, AD is the common chord of the circles on AB and AC as diameters. Similarly, the other two altitudes BE and CF are the other two common chords. Thus, the point of intersection of the common chords is the point of intersection of AD, BE and CF i.e. the orthocentre of `DeltaABC`. |
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| 169. |
Let `P(alpha,beta)` be a point in the first quadrant. Circles are drawn through P touching the coordinate axes. Equation of common chord of two circles isA. `x +y = alpha - beta`B. `x +y = 2 sqrt(alpha beta)`C. `x +y = alpha +beta`D. `alpha^(2) -beta^(2) = 4 alpha beta` |
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Answer» Correct Answer - C `S_(1) = x^(2) +y^(2) - 2r_(i)(x+y) +r_(i)^(2) =0`, where `i =1,2` Thus, equation of common chord. `2(r_(2)-r_(1)) (r+y) +r_(1)^(2) -r_(2)^(2) =0` `rArr 2(x+y) = (r_(1)+r_(2)) = 2 (alpha + beta)` `rArr x +y = alpha + beta`. |
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| 170. |
If the chord of contact of the tangents drawn from the point `(h , k)`to the circle `x^2+y^2=a^2`subtends a right angle at the center, then prove that `h^2+k^2=2a^2dot`A. `h^(2)+k^(2)=a^(2)`B. `2(h^(2)+k^(2))=a^(2)`C. `h^(2)-k^(2)=a^(2)`D. `h^(2)+k^(2)=2a^(2)` |
| Answer» Correct Answer - D | |
| 171. |
Statement-1: The line `x+9y-12=0` is the chord of contact of tangents drawn from a point P to the circle `2x^(2)+2y^(2)-3x+5y-7=0`. Statement-2: The line segment joining the points of contacts of the tangents drawn from an external point P to a circle is the chord of contact of tangents drawn from P with respect to the given circleA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - D Statement-2 is the definition of chord of contact of tangents drawn from a point to a given circle. So, it is true. Let x+9y-12=0 be the chord of contact of tangents drawn from a point p(h, k) to the circle `2x^(2)+2y^(2)-3x+5y-7=0`. Then its equation is `hx+ky-(3)/(4)(x+h)+(5)/(4)(y+k)-(7)/(2)=0` or, `(4h-3)x+(4k+5)y-3h+5k-14=0` This should be same as `x+9y-12=0`. `:. (4h-3)/(1)=(4k+5)/(9)=(3h-5k+14)/(12) rArr h=1, k=1` But, (1,1) lies inside hte circle `2x^(2)+2y^(2)-3x+5y-7=0`. So, `x+9y-12=0` cannot be the chord of contact of tangents drawn from P. Hence, statement-1 is not true. |
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| 172. |
Find the condition if the circle whose equations are `x^2+y^2+c^2=2a x`and `x^2+y^2+c^2-2b y=0`touch one another externally.A. `(1)/(b^(2))+(1)/(c^(2))=(1)/(a^(2))`B. `(1)/(c^(2))+(1)/(a^(2))=(1)/(b^(2))`C. `(1)/(a^(2))+(1)/(b^(2))=(1)/(c^(2))`D. none of these |
| Answer» Correct Answer - C | |
| 173. |
Find the locus of the midpoint of the chord of the circle `x^2+y^2-2x-2y=0`, which makes an angle of `120^0`at the center.A. `x^(2)+y^(2)-2x-2y+1=0`B. `x^(2)+y^(2)+x+y-1=0`C. `x^(2)+y^(2)-2x-2y-1=0`D. none of these |
| Answer» Correct Answer - A | |
| 174. |
The equation of the circle described on the common chord of the circles `x^(2)+y^(2)+2x=0` and `x^(2)+y^(2)+2y=0` as diameter, isA. `x^(2)+y^(2)+x-y=0`B. `x^(2)+y^(2)-x-y=0`C. `x^(2)+y^(2)-x+y=0`D. `x^(2)+y^(2)+x+y=0` |
| Answer» Correct Answer - D | |
| 175. |
IF `(alpha, beta)` is a point on the chord PQ of the circle `x^(2)+y^(2)=25`, where the coordinates of P and Q are (3, -4) and (4, 3) respectively, thenA. `3le alpha le 4 and -4 le beta le 3`B. `-4le alpha le 3 and 3 le beta le 4`C. `alpha 3 and -4 le beta le 4`D. none of these |
| Answer» Correct Answer - A | |
| 176. |
The angle of intersection of the circles `x^(2)+y^(2)=4` and `x^(2)+y^(2)+2x+2y`, isA. `pi//2`B. `pi//3`C. `pi//6`D. `pi//4` |
| Answer» Correct Answer - D | |
| 177. |
If `theta`is the angle between the two radii (one to each circle) drawn from oneof the point of intersection of two circles `x^2+y^2=a^2`and `(x-c)^2+y^2=b^2,`then prove that the length of the common chord of the two circles is `(2a bsintheta)/(sqrt(a^2+b^2-2a bcostheta))`A. `(ab)/(sqrt(a^(2)+b^(2)-2ab cos theta))`B. `(2ab)/(sqrt(a^(2)+b^(2)-2ab cos theta))`C. `(2ab sin theta)/(sqrt(a^(2)+b^(2)-2ab cos theta))`D. `(2ab cos theta)/(sqrt(a^(2)+b^(2)-2ab cos theta))` |
| Answer» Correct Answer - C | |
| 178. |
The range of values of a for which the point (a, 4) is outside the circles `x^(2)+y^(2)+10x=0` and `x^(2)+y^(2)-12x+20=0`, isA. `(-oo, -8) uu(-2,6) uu (6, oo)`B. (-8, -2)C. `(-oo, -2)uu(-2, oo)`D. none of these |
| Answer» Correct Answer - A | |
| 179. |
The range of g so that we have always a chord of contact of tangents drawn from a real point `(alpha, alpha)` to the circle `x^(2)+y^(2)+2gx+4y+2=0`, isA. (-3, 0)B. (-4, 1)C. (-4, 0)D. none of these |
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Answer» Correct Answer - C For a real chord of contact of tangents drawn from `(alpha, alpha)` to the circle `x^(2)+y^(2)=2gx+4y+2=0`, the point `(alpha, alpha)` must lie outside the circle. `:. alpha^(2)+alpha^(2)+2g alpha + 4 alpha + 2 gt 0` `rArr alpha^(2)+g alpha + (2 alpha + 1) gt 0` `rArr alpha^(2)+(g+2)alpha+1gt0` Since `alpha` assumes real values . Therefore, `(g+2)^(2)-4 gt = 0` `rArr g^(2)+4h lt 0 rArr -4 lt g lt 0 rArr g in (-4, 0)`. |
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| 180. |
The chords of contact of the pair of tangents drawn from each point on the line `2x + y=4` to the circle `x^2 + y^2=1` pass through the point (a,b) then 4(a+b) isA. (1/2, 1/4)B. (1/4, 1/2)C. (1, 1/2)D. (1/2, 1) |
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Answer» Correct Answer - A Let P(t, 4-2t) be any point of the line 2x+y=4. The equation of the chord of contact of tangents drawn from P to the circle `x^(2)+y^(2)=1` is `tx+(4-2t)y=1rArr(4y-1)+t(x-2y)=0` Clearly, it passes through the point of intersection of the lines 4y-1=0 and x-2y=0 i.e. (1/2, 1/4). |
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| 181. |
The distance between the chords of contact of tangents to the circle `x^2+y^2+2gx +2fy+c=0` from the origin & the point (g,f) isA. `g^(2)+f^(2)`B. `(1)/(2)(g^(2)+f^(2)+c)`C. `(g^(2)+f^(2)+c)/(2sqrt(g^(2)+f^(2)))`D. `(g^(2)+f^(2)-c)/(2sqrt(g^(2)+f^(2)))` |
| Answer» Correct Answer - D | |
| 182. |
All chords.of the curve `x^2+y^2-10x-4y+4=0` which make a right angle at (8,-2) pass throughA. `(2,5)`B. `(-2,-5)`C. `(-5,-2)`D. `(5,2)` |
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Answer» Correct Answer - D `(8,-2)` lies on the circle `(x-5)^(2) +(y-2)^(2) =25` and a chord making a right angle at `(8,-2)` must be a diameter of the circle. So they all pass through the centre `(5,2)`. |
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| 183. |
PQ is a chord of the circle `x^(2)+y^(2)-2x-8=0` whose mid-point is (2, 2). The circle passing through P, Q and (1, 2) isA. `x^(2)+y^(2)-7x+10y+28=0`B. `x^(2)+y^(2)-7x-10y+22=0`C. `x^(2)+y^(2)-7x-10y+22=0`D. `x^(2)+y^(2)+7x+10y-22=0` |
| Answer» Correct Answer - B | |
| 184. |
the length of the chord of the circle `x^(2)+y^(2)=25` passing through (5, 0) and perpendicular to the line `x+y=0`, isA. `5 sqrt(2)`B. `5sqrt(2)`C. `2sqrt(5)`D. none of these |
| Answer» Correct Answer - A | |
| 185. |
If the angle of intersection of the circle `x^2+y^2+x+y=0`and `x^2+y^2+x-y=0`is `theta`, then the equation of the line passing through (1, 2) and making anangle `theta`with the y-axis is`x=1`(b) `y=2``x+y=3`(d) `x-y=3`A. `pi//6`B. `pi//4`C. `pi//3`D. `pi//2` |
| Answer» Correct Answer - D | |
| 186. |
Statement 1 : The equation `x^2+y^2-2x-2a y-8=0`represents, for different values of `a ,`a system of circles passing through two fixed points lying on thex-axis.Statement 2 : `S=0`is a circle and `L=0`is a straight line. Then `S+lambdaL=0`represents the family of circles passing through the points ofintersection of the circle and the straight line (where `lambda`is an arbitrary parameter).A. `x^(2)+y^(2)-2y=0`B. `x^(2)+y^(2)-2x-8=0`C. `x^(2)+y^(2)-2y=8`D. `x^(2)+y^(2)-2x-2y=8` |
| Answer» Correct Answer - B | |
| 187. |
From points on the straight line 3x-4y + 12 = 0, tangents are drawn to the circle `x^2 +y^2 = 4`. Then, the chords of contact pass through a fixed point. The slope of the chord of the circle having this fixed point as its mid-point isA. `(4)/(3)`B. `(1)/(2)`C. `(1)/(3)`D. none of these |
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Answer» Correct Answer - D Any point on the given line can be taken as `(alpha, (3alpha+12)/(4))` Then the equation of the chord of contact of tangents from this point to `x^(2) +y^(2) = 4` is `x alpha +y ((3alpha +12))/(4) - 4 =0` or `alpha (4x +3y) +(12 y -16) =0` For different values of `alpha`, the above chord passes through `P(-1,(4)/(3))` which is the point of intersection of `4x +3y = 0` and `3y -4 =0`. Now slope of line joining center of circle `O(0,0)` and point `P(-1,(4)/(3))` is `-4//3`. Thus, slope of the chord for which P is mid-point is `3//4`. |
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| 188. |
Let `L_1` be a straight line passing through the origin and ` L_2` be the straight line `x + y = 1` if the intercepts made by the circle `x^2 + y^2-x+ 3y = 0` on `L_1` and `L_2` are equal, then which of the following equations can represent `L_1`?A. `x+y=0, x-7y=0`B. `x-y=0, x+7y=0`C. `7x+y=0`D. `x-7y=0` |
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Answer» Correct Answer - B The centre and radius of the given circle are `((1)/(2), -(3)/(2))` and `sqrt((5)/(2))` respectively. Let `y=mx` be the equation of `L_(1)`. Then, `p_(1)`= Length of the intercept on `L_(1)` `rArr p_(1)=2sqrt((sqrt((5)/(2)))^(2)-((m+3)/(2sqrt(m^(2)+1)))^(2))=2sqrt((5)/(2)-((m+3)^(2))/(4(m^(2)+1)))` and, `rArr p_(2)=` Length of the intercept on `L_(2)` `rArr p_(2)=2sqrt((sqrt((5)/(2)))^(2)-(sqrt(2))^(2))=2sqrt((5)/(2)-2)=sqrt(2)` Now, `p_(1)=p_(2)` `rArr 2sqrt((5)/(2)-((m+3)^(2))/(4(m^(2)+1)))=sqrt(2)` `rArr 5-((m+3)^(2))/(4(m^(2)+1))=1` `rArr 7m^(2)-6m-1=0` `rArr (m-1)(7m+1)=0rArr m=1, -(1)/(7)` So, the equations of `L_(1)` are `y=x` and `7y=-x`. |
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| 189. |
If the chord of contact of tangents from a point `P`to a given circle passes through `Q ,`then the circle on `P Q`as diameter.cuts the given circle orthogonallytouches the given circle externallytouches the given circle internallynone of theseA. cuts the given circle orthogonallyB. touches the given circle externallyC. touches the given circle internallyD. none of these |
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Answer» Correct Answer - A Let `P(x_(1), y_(1))` and `Q(x_(2), y_(2))` be the given points and `x^(2)+y^(2)=a^(2)` be the circle. The chord of contact of tangents drawn from `P(x_(1), y_(1))` to `x^(2)+y^(2)=a^(2)` is `x x_(1)+y y_(1)=a^(2)` If it passes through `Q(x_(2), y_(2))`, then `x_(1) x_(2)+y_(1)y_(2)=a^(2) " " ...(i)` The equation of the circle on PQ as diameter is `(x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0` `rArr x^(2)+y^(2)-x(x_(1)+x_(2))-y(y_(1)+y_(2))+x_(1)x_(2)+y_(1)y_(2)=0` This circle will cut the given circle orthogonally, if `0(x_(1)+x_(2))+0(y_(1)+y_(2))=a^(2)+x_(1)x_(2)+y_(1)y_(2)` `rArr x_(1)x_(2)+y_(1)y_(2)=a^(2)=0`, which is true. [Using (i)] |
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| 190. |
For all values of `m in R` the line `y - mx + m - 1 = 0` cuts the circle `x^2 + y^2 - 2x - 2y + 1 = 0` at an angleA. `(pi)/(3)`B. `(pi)/(6)`C. `(pi)/(2)`D. `(pi)/(4)` |
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Answer» Correct Answer - C The given line is a diameter to the circle. |
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| 191. |
The area of the circle centred at (1,2) and passing through (4,6) isA. `5pi`B. `10 pi`C. `25 pi`D. none of these |
| Answer» Correct Answer - C | |
| 192. |
If the line `y = mx - (m-1)` cuts the circle `x^2+y^2=4` at two real and distinct points thenA. `m in (1, 2)`B. `m=1`C. `m=2`D. `m in R` |
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Answer» Correct Answer - D If the line `y= mx-(m-1)` cuts the circle in two distinct points, then Length of the perpendicular from the centre `lt` Radius. `rArr |(mxx0-0-(m-1))/(sqrt(m^(2)+1))| lt 2` `rArr (|m-1|)/(sqrt(m^(2)+1))lt2` `rArr (m-1)^(2)lt 2(m^(2)+1)` `rArr m^(2)+2m+1 gt 0` `rArr (m+1)^(2)gt0`, which is true for all `m in R`. Hence, option (d) is correct. `ul("ALITER")` The equation of the line is `y-1=m(-1)`. Clearly, it passes through (1, 1) which is an interior point of the circle. So, the line cuts the circle for all values of m. |
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| 193. |
The line `y = mx` intersects the circle `x^(2)+y^(2) -2x - 2y = 0` and `x^(2)+y^(2) +6x - 8y =0` at point A and B (points being other than origin). The range of m such that origin divides AB internally isA. `-1 lt m lt (3)/(4)`B. `m gt (4)/(3)` or `m lt -2`C. `-2 lt m lt (4)/(3)`D. `m gt -1` |
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Answer» Correct Answer - A The tangents at the origin to `C_(1)` and `C_(2)` are `x +y =0, 3x -4y =0`, respectively. Slope of the tangents are `-1` and `(3)/(4)`, respectively. Then if `-1 lt m lt (3)/(4)`, then origin divides AB internally. |
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| 194. |
Three sided of a triangle have equations `L_1-=y-m_i x=o; i=1,2a n d3.`Then `L_1L_2+lambdaL_2L_3+muL_3L_1=0`where `lambda!=0,mu!=0,`is the equation of the circumcircle of the triangle if`1+lambda+mu=m_1m_2+lambdam_2m_3+lambdam_3m_1``m_1(1+mu)+m_2(1+lambda)+m_3(mu+lambda)=0``1/(m_3)+1/(m_1)+1/(m_1)=1+lambda+mu`none of theseA. `lambda(m_(2)+m_(3))+mu (m_(3)+m_(1))+v(m_(1)+m_(2))=0`B. `lambda(m_(2)m_(3)-1)+mu(m_(3)m_(1)-1)+v(m_(1)m_(2)-1)=0`C. both (a) and (b) hold togetherD. none of these |
| Answer» Correct Answer - C | |
| 195. |
If the circle `x^2 + y^2 = a^2` cuts off a chord of length `2b` from the line `y = mx +c`, thenA. `sqrt(a^(2)(a+m^(2))) lt c`B. `sqrt(a^(2)(1-m^(2)))ltc`C. `sqrt(a^(2)(a+m^(2))) gt c`D. `sqrt(a^(2)(1-m^(2)))gtc` |
| Answer» Correct Answer - C | |
| 196. |
if `y = mx` is a chord of a circle of radius `a` and the diameter of the circle lies along `x`-axis and one end of this chord in origin .The equation of the circle described on this chord as diameter isA. `(1+m^(2))(x^(2)+y^(2))-2a(x+my)=0`B. `(1-m^(2))(x^(2)+y^(2))-2a(x+my)=0`C. `(1+m^(2))(x^(2)+y^(2))+2a(x+my)=0`D. none of these |
| Answer» Correct Answer - A | |
| 197. |
If the `y=mx+1,` of the circle `x^2+y^2=1` subtends an angle of measure `45^@` of the major segment of the circle then value of `m` is -A. 2B. -2C. 1D. none of these |
| Answer» Correct Answer - C | |
| 198. |
Angle subtended by an arc at its centre is ‘x°’ then, angle subtended by it at any other point on the circle isA) x°/2B) 2x° C) x° D) 180 – x° |
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Answer» Correct option is (A) \(\frac{x^\circ}{2}\) Angle subtended by an arc at the centre is double then the angle subtended by that arc at any other point on the circumference of the circle. \(\therefore\) \(x^\circ\) = 2 \(\times\) Angle subtended by arc at any other point on the circle. \(\therefore\) Required angle = \(\frac{x^\circ}{2}\) Correct option is A) x°/2 |
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| 199. |
If the line y=mx +1 meets the circle `x^(2)+y^(2)+3x=0` in two points equidistant and on opposite sides of x-axis, thenA. 3m-2=0B. 2m+3=0C. 3m+2=0D. 2m-3=0 |
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Answer» Correct Answer - B If the line y=mx+1 meets the circle `x^(2)+y^(2)+3x=0` in two points equidistant and on opposite sides of x-axis then the equation `x^(2)+(mx+1)^(2) +3x=0` or `x^(2)(1+m^(2))+(2m+3)x+1=0` must have roots equal in magnitude but opposite in sign. `:. `Sum of the roots `=0 rArr - (2m+3)/(1+m)=0rArr 2m+3=0` |
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| 200. |
In figure, `angleADC=130^(@)" and chord BC=chord BE. Find "angleCBE`. |
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Answer» We have, `angleADC=130^(@)` and chord BC=chord BE. Suppose, we consider the points A, B, C and D form a cyclic quadrilateral. Since, the sum of opposite angles of a cyclic quadrilateral ADCB is ` 180^(@)`. `:. angleADC+angleOBC=180^(@)` `rArr 130^(@)+angleOBC=180^(@)` `rArr angleOBC=180^(@)-130^(@)=50^(@)` `"In" DeltaBOC and DeltaBOE`, BC=BE [given equal chord] OC=OE [both are the radius of a circle] and OB=OB [common side] `:. DeltaBOC cong DeltaBOE` [by SSS songruence rule] `rArr angle OBC=angleOBE=50^(@)` [by CPCT] Now, `angleCBE=angleCBO+angleEBO` `=50^(@)+50^(@)=100^(@)` |
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