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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.
| 101. |
Find the equation of the circle of radius 5 cm, whose centre lies on the lies on the y-axis and which passes through the point ` (3, 2 )` |
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Answer» Correct Answer - ` (x ^(2) + y ^(2) - 12y + 11 =0 ) or (x ^(2) + y ^(2) + 4y - 21=0 )` Let the centre of the circle be `C( 0, k)` It is given that the circle passes through ` P(3, 2)` Then `, |CP|^(2) = 5^(2) hArrf (0 - 3) ^(2) + ( 2 - k ) ^(2) = 5^(2)` `" " hArr ( 2- k ) = pm 4 hArr k = 6 or k = - 2 ` ` therefore ` the centre of the circles is ` (0, 6) or (0, - 2 )` Hence, the equation of the circle is ` " "[(x - 0) ^(2) + ( y - 6) ^(2) = 5^(2) ] or [ ( x - 0 ) ^(2) + (y + 2 ) ^(2) = 5 ^(2) ] `. |
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| 102. |
Show that the equation x2 + y2 – 4x + 6y – 5 = 0 represents a circle. Find its centre and radius. |
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Answer» The general equation of a conic is as follows ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 where a, b, c, f, g, h are constants For a circle, a = b and h = 0. The equation becomes: x2 + y2 + 2gx + 2fy + c = 0…(i) Given, x2 + y2 – 4x + 6y – 5 = 0 Comparing with (i) we see that the equation represents a circle with 2g = - 4 ⇒ g = - 2, 2f = 6 ⇒ f = 3 and c = - 5. Centre ( - g, - f) = { - ( - 2), - 3} = (2, - 3). Radius = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{(-2)^2+3^2-(-5)}\) = \(\sqrt{4+9+5}\) = \(\sqrt{18}\) = \(3\sqrt{2}\) |
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| 103. |
Find the equation of a circle, the end points of one of whose diameters are ` A (2, -3) and B(-3, 5)`. |
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Answer» We know that the equation of a circle, the end points of one of whose diameters are `(x_1, y_1) and (x_2, y_2) `, is given by ` " "(x - x_1) (x - x _2 ) + (y -y_1 ) (y - y_2) = 0 ` Here, ` x_1 = 2, y_1 = - 3 and x_2 = - 3, y_2 = 5` ` therefore ` the required equation of the circle is ` (x - 2 ) (x + 3) + ( y + 3 ) ( y - 5)=0` ` rArr x ^(2) + y ^(2) + x - 2y - 21 = 0` |
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| 104. |
Find the equation of the circle, the coordinates of the end points of one of whose diameters are |
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Answer» The equation of a circle passing through the coordinates of the end points of diameters is: (x - x1) (x - x2) + (y - y1)(y - y2) = 0 Substituting, values:(x1, y1) = (3, 2) & (x2, y2) = (2, 5) We get: (x - 3)(x - 2) + (y - 2)(y - 5) = 0 ⇒ x2 - 2x - 3x + 6 + y2 - 5y - 2y + 10 = 0 ⇒ x2 + y2 - 5x - 7y + 16 = 0 x2 + y2 - 5x - 7y + 16 = 0 |
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| 105. |
Find the equation of the circle, the coordinates of the end points of one of whose diameters are A(5, - 3) and B(2, - 4) |
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Answer» The equation of a circle passing through the coordinates of the end points of diameters is: (x - x1) (x - x2) + (y - y1)(y - y2) = 0 Substituting, values:(x1, y1) = (5, - 3) & (x2, y2) = (2, - 4) We get: (x - 5)(x - 2) + (y + 3)(y + 4) = 0 ⇒ x2 - 2x - 5x + 10 + y2 + 3y + 4y + 12 = 0 ⇒ x2 + y2 - 7x + 7y + 22 = 0 x2 + y2 - 7x + 7y + 22 = 0 |
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| 106. |
The sides of a rectangle are given by the equations x=-2, x = 4, y=-2 andy=5. Find the equation of the circle drawn on the diagonal of this rectangle as its diameter. |
| Answer» Correct Answer - ` x ^(2) + y ^(2) - 2x - 3y - 18 =0` | |
| 107. |
The equation of the circle passing through the point of intersection ofthe circles `x^2+y^2-4x-2y=8`and `x^2+y^2-2x-4y=8`and the point `(-1,4)`is`x^2+y^2+4x+4y-8=0``x^2+y^2-3x+4y+8=0``x^2+y^2+x+y=0``x^2+y^2-3x-3y-8=0`A. `x^(2)+y^(2)+4x+4y-8=0`B. `x^(2)+y^(2)-3x+4y+8=0`C. `x^(2)+y^(2)+x+y-8=0`D. `x^(2)+y^(2)-3x-3y-8=0` |
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Answer» Correct Answer - 4 The equation of any circle through the points of intersection of the given circles is `x^(2)+y^(2)-4x-2y-8+k(x^(2)+y^(2)-2x-4y-8)=0` (1) Since circle (1) passes through `(-1,4)`, we have `k=1` Therefore, the requiredcircle is `x^(2)+y^(2)-3x-3y-8=0` |
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| 108. |
If P and Q are the points of intersection of the circles `x^2+y^2+3x+7y+2p=0` and `x^2+y^2+2x+2y-p^2=0` then there is a circle passing through P,Q and (1,1) forA. all values of pB. all except one value of pC. all except two values of pD. exactly one value of p |
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Answer» Correct Answer - 2 `x^(2)+y^(2)+3x+7y+2p-5+lambda(x^(2)+y^(2)+2x+2y-p^(2))=0`, `lambda cancel(=) -1`, passes through point of intersection of given circles. Sincet it passes through `(1,1)` `7-2p+lambda(6-p^(2))=0` `implies 7-2p+6lambda -lambdap^(2)=0` If `lambda = -1`, then `7-2p-6+p^(2)=0` `implies p^(2)-2p+1=0` `:. p =1` If `lambda cancel (=)-1` then `pcancel(=) 1`. Therefore, all values of p are possible except `p=1` |
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| 109. |
Show that the equation of the circle passing through (1, 1) and thepoints of intersection of the circles `x^2+y^2+13 x-13 y=0`and `2x^2+2y^2+4x-7y-25=0`is `4x^2+4y^2+30 x-13 y-25=0.`A. `4x^(2)+4y^(2)-30x-10y=25`B. `4x^(2)+4y^(2)+30x-13y-25=0`C. `4x^(2)+4y^(2)+17x-10y-25=0`D. None of the above |
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Answer» Correct Answer - B The required equation of circle is `(x^(2)+y^(2)+13x-3y)+lambda(11x+(1)/(2)y+(25)/(2))=0" "...(i)` Its passing through (1, 1), `rArr 12 + lambda(24)=0` `rArr lambda = (1)/(2)` On putting in Eq. (i) , we get `x^(2)+y^(2)+13x-3y-(11)/(2)x-(1)/(4)y-(25)/(4)=0` `rArr 4x^(2)+4y^(2)+52x-12y-22x-y-25=0` `rArr 4x^(2)+4y^(2)+30x-13y-25=0` |
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| 110. |
Find the equation of the circle, the coordinates of the end points of one of whose diameters are (i) ` A (3, 2 ) and B(2, 5)` (ii) `A (5, - 3) and B(2, - 4 )` (iii) ` A(-2, -3) and B(-3, 5)` (iv) `A(p, q), B (r, s )` |
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Answer» Correct Answer - (i) ` x ^(2) + y ^(2) - 5x - 7y + 16 = 0` (ii) ` x ^(2 ) + y ^(2) - 7x + 7y + 22 = 0` (iii) ` x ^(2) + y ^(2) + 5x - 2y - 9 = 0` (iv) ` (x - p) (x - r) + (y - q ) (y - s ) = 0 ` |
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| 111. |
Find the equation of the circle passing through the point `(-1,3)` and having its centre at the point of intersection of the lines `x-2y=4` and `2x+5y+1=0` |
| Answer» Correct Answer - ` x ^(2) + y ^(2) - 4x + 2y- 20 = 0` | |
| 112. |
Find the equation of the circle which passes through the vertices of the triangle formed by `L_(1) = x + y + 1=0` `L_(2) = 3x + y + 5=0and L_(3) = 2x + y - 5 =0`A.B.C.D. |
| Answer» Correct Answer - i.e.,`x^(2)+y^(2)-30x-10y+25=0` | |
| 113. |
Find the equation of th circle whose centre is ` (2, -3)` and which passes through the intersection of the lines ` 3x + 2y = 11 and 2x + 3y = 4`. |
| Answer» Correct Answer - ` x ^(2) + y ^(2) - 4x + 6y + 3 = 0` | |
| 114. |
Find the equation of the circle which touches the axes and whose centre lies on x – 2y = 3. |
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Answer» Let us assume the circle touches the axes at (a, 0) and (0, a) and we get the radius to be |a|. We get the centre of the circle as (a, a). This point lies on the line x – 2y = 3 a – 2(a) = 3 -a = 3 a = – 3 Centre = (a, a) = (-3, -3) and radius of the circle(r) = |-3| = 3 We have circle with centre (-3, -3) and having radius 3. We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2 Now by substituting the values in the equation, we get (x – (-3))2 + (y – (-3))2 = 32 (x + 3)2 + (y + 3)2 = 9 x2 + 6x + 9 + y2 + 6y + 9 = 9 x2 + y2 + 6x + 6y + 9 = 0 ∴ The equation of the circle is x2 + y2 + 6x + 6y + 9 = 0. |
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| 115. |
A circle C whose radius is 1 unit, touches the x-axis at point A. The centre Q of C lies in first quadrant. The tangent from origin O to the circie touches it at T and a point P lies on it such that `DeltaOAP` is a right angled triangle at A and its perimeter is 8 units. The length of `QP` isA. `(x-2)^(2)+(y-1)^(2)=1`B. `{x-(sqrt(3)-sqrt(2))}^(2)+(y-1)^(2)=1`C. `(x-sqrt(3))^(2)+(y-1)^(2)=1`D. none of these |
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Answer» Correct Answer - 1 Now,` OA=2.` So, center of the circle is `Q (2,1)` Equation of circles is `( x-2)^(2)+ (y -1)^(2)=1` |
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| 116. |
The length of the diameter of the circle which touches the x-axis atthe point (1, 0) and passes through the point (2, 3) is(1) `(10)/3`(2) `3/5`(3) `6/5`(4) `5/3`A. `(10)/(3)`B. `(3)/(5)`C. `(6)/(5)`D. `(5)/(3)` |
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Answer» Correct Answer - A |
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| 117. |
A circle C whose radius is 1 unit, touches the x-axis at point A. The centre Q of C lies in first quadrant. The tangent from origin O to the circie touches it at T and a point P lies on it such that `DeltaOAP` is a right angled triangle at A and its perimeter is 8 units. The length of `QP` isA. `3y=4x`B. `x-sqrt(2)y=0`C. `y-sqrt(3)x=0`D. none of these |
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Answer» Correct Answer - 1 `AP = AQ+PQ= 1+5//3=8//3` So, slope of OP is `4//3` Hence, equation of OP is `4x-3y=0`. |
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| 118. |
A circle passes through the points (-1,3) and (5,11) and its radius is 5. Then, its centre isA. (-5,0)B. (-5,7)C. (2,7)D. (5,0) |
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Answer» Correct Answer - C |
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| 119. |
A circle passes through point `(3, sqrt(7/2))` and touches the line-pair `x^2 - y^2 - 2x +1 = 0`. Centre of circle lies inside the circle `x^2 + y^2 - 8x + 10y + 15 = 0`. Coordinates of centre of circle are given by (A) `(4,0)` (B) `(5,0)` (C) `(6,0)` (D) `(0,4)`A. (4,0)B. (4,2)C. (6,0)D. (7,9) |
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Answer» Correct Answer - A::C |
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| 120. |
Show that the equation `3x^2 + 3y^2 + 6x - 4y -1 = 0` represents a circle Find its centre and radius. |
| Answer» Correct Answer - Centre ` (- 1, (2)/(3))`, radius ` = (4)/(3)` | |
| 121. |
Show that the equaion ` x ^(2) + y ^(2) - 6x + 4y - 36 = 0 ` represents a circle. Also, find its centre and radius. |
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Answer» The given equation is ` x ^(2) + y ^(2) - 6x + 4y - 36 = 0` . This is of the form ` x ^(2) + y^(2) + 2gx + 2 fy + c= 0`, `" "` where ` 2g = - 6, 2f = 4 and c = - 36` ` therefore g = -3, f = 2 and c = - 36`. Hence, the given equation represents a circle. Centre of the circle `= (-g, -f ) = (3, - 2 )` Radius of the circle = ` sqrt ( g ^(2) + f ^(2) - c ) = sqrt (3 ^(2) + (-2 ) ^(2) + 36 ) = sqrt ( 49) = 7 ` units. |
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| 122. |
Show that the equation 3x2 + 3y2 + 6x - 4y – 1 = 0 represents a circle. Find its centre and radius. |
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Answer» The general equation of a conic is as follows ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 where a, b, c, f, g, h are constants For a circle, a = b and h = 0. The equation becomes: x2 + y2 + 2gx + 2fy + c = 0…(i) Given, 3x2 + 3y2 + 6x - 4y - 1 ⇒ x2 + y2 + 2x - \(\frac{4}{3}\)y - \(\frac{1}{3}\) = 0 Comparing with (i) we see that the equation represents a circle with 2g = 2 ⇒ g = 1.2f = -\(\frac{4}{3}\) ⇒ f = -\(\frac{2}{3}\) and c = -\(\frac{1}{3}\) Centre (-g,-f) = {-1,-(-\(\frac{2}{3}\))} = \(\Big(-1,\frac{2}{3}\Big)\) Radius = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{1^2+(-\frac{2}{3})^2-(-\frac{1}{3})}\) = \(\sqrt{1+\frac{4}{9}+\frac{1}{3}}\) = \(\sqrt{\frac{16}{9}}\) = \(\frac{4}{3}\) |
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| 123. |
Find the equations of the tangents to the circle `x^2+y^2-6x+4y=12`which are parallel to the straight line`4x+3y+5=0` |
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Answer» Given circle is `x^(2)+y^(2)-6x+4y-12=0` Centre of the circle is `(3,-2)` and radius is `sqrt((-3)^(2)+(2)^(2)-(-12))=5` Given that tangents are parallel to the line `4x+3y+5=0`. Therefore, the equations of tangents take the form `4x+3y+c=0`. Distance of centre of the circle from the tangent line is radius. `:. |(4(3)+3(-2)+k)/(sqrt(16+9))|=5` `implies 6+k=+-25` `implies k=19` and `-31` Therefore, the equations of required tangents are `4x+3y+19=0` and `4x+3y-31=0` Alternative method `:` Given circle in center-radius form form is `(x-3)^(2)+(y+2)^(2)=5^(2)` Tangents are parallel to the line `4x+3y+5=0` . `:. ` Slope of tangents `=-(4)/(3)` Hence, using the equation of tangents `y+f=m(x+g)+-sqrt(m^(2)+1)sqrt(g^(2)+f^(2)-c)`, required equations of tangents are obtained as `y+2=-(4)/(3)(x-3)+-5sqrt((-(4)/(3))^(2)+1)` or `4x+3y+19=0` and `4x+3y-31=0` |
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| 124. |
Find the equation of the ciracle whose centre is `(-1,2)` and which passes through `(5,6)`A.B.C.D. |
| Answer» Correct Answer - `x^(2) + y^(2) + 2x-4y-47 = 0` | |
| 125. |
Find the centre and radius of the circle `3x^(2)+ 3y^(2) - 6x + 4y - 4 = 0`A.B.C.D. |
| Answer» Correct Answer - `= 5/3` | |
| 126. |
Find the centre and radius of the circle `x^(2) + y^(2) + 2x-4y-4=0`A.B.C.D. |
| Answer» Correct Answer - `=3` | |
| 127. |
In figure, if ∠ABC 45°, then prove that OA ⊥ OC. |
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Answer» ∠ABC = 45° (given) We know that angle subtended by an arc of a circle at center is double the angle subtended at remaining part of circle by same arc. Then ∠AOC = 2∠ABC = 2 × 45° = 90° Thus, OA ⊥ OC |
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| 128. |
Degree measure of major arc is :(A) less than 180°(B) more than 180°(C) 270°(D) 360° |
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Answer» Answer is (B) more than 180° |
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| 129. |
Sum of opposite angles in a cyclic quadrilateral is :(A) 270°(B) 360°(C) 180°(D) 90° |
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Answer» Answer is (C) 180° |
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| 130. |
In fig, if O is center of circle, then ∠AOB will be :(A) 70°(B) 110°(C) 120°(D) 140° |
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Answer» Answer is (D) 140° Join CO In ∆AOC, AO = CO (Radius of same circle) ⇒ ∠OAC = ∠OCA ∴ ∠OCA = 30° In ∆OCA ∠AOC + ∠OCA + ∠OAC = 180° [∵ Sum of all the angles of triangle is 180°] ⇒ ∠AOC + 30° + 30° = 180° ⇒ ∠AOC + 60° = 180° ∴ ∠AOC = 180° – 60° = 120° Similarly, OB = OC ∠OBC = ∠OCB = 40° In ∆OBC ∠BOC + ∠OBC + ∠OCB = 180° ⇒ ∠BOC + 40° + 40° = 180° ∴ ∠BOC = 180° – 80° = 100° ∴ ∠AOB = 360° – (∠AOC + ∠BOC) = 360° – (120° + 100°) = 360° – 220° ∠AOB = 140° |
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| 131. |
In figure, ∠BAC will be :(A) 80°(B) 160°(C) 90°(D) 200° |
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Answer» Answer is (A) 80° We know that sum of angles at center is 360°. ∠BOC – ∠BOA + ∠COA = 360° ⇒ ∠BOC + 85° + 115° = 360° ⇒ ∠BOC + 200° = 360° ∴ ∠BOC = 360° – 200° = 160° Angle subtended at center of a circle is double the angle at remaining part by same arc. ∠BOC = 2∠BAC ⇒ ∠BOC = \(\frac { \angle BOC }{ 2 } \) = \(\frac { { 160 }^{ \circ } }{ 2 } \) ∴ ∠BAC = 80° |
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| 132. |
If `alpha` is the angle subtended at `P(x_(1),y_(1))` by the circle `S-=x^(2)+y^(2)+2gx+2fy+c=0` thenA. `cotalpha=(sqrtS_(1))/(sqrt((g^(2)+f^(2)-c)))`B. `cot""(alpha)/(2)=(sqrtS_(1))/(sqrt((g^(2)+f^(2)-c)))`C. `tanalpha=(2sqrt((g^(2)+f^(2)-c)))/(sqrtS_(1))`D. `alpha=2tan^(-1)((sqrt((g^(2)+f^(2)-c)))/(sqrtS_(1)))` |
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Answer» Correct Answer - B::D |
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| 133. |
Find the length of the tangent drawn from any point on the circle `x^2+y^2+2gx+2fy+c_1=0`to the circle `x^2+y^2+2gx+2fy+c_2=0` |
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Answer» Correct Answer - `sqrt(c_(2)-c_(1))` Let `(x_(1),y_(1))` be any point on the circle `x^(2)+y^(2)+2gx+2fy+c_(1)=0` `:. x_(1)^(2)+y_(1)^(2)+2gx_(1)+2fy_(1)+c_(1)=0` (1) The length of the tangent from `(x_(1),y_(1))` to the circle `x^(2)+y^(2)+2gx+2fy+c_(2)=0` is `sqrt(x_(1)^(2)+y_(1)^(2)+2gx_(1)+2fy_(1)+c_(2))=sqrt(c_(2)-c_(1))` [Using (1)] |
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| 134. |
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) 4A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C PLAN Number of common tangents depend on the position of the circle with respect to each other. (i) If circles touch internally `rArr C_(1)C_(2)=r_(1)+r_(2)`, 3 common tangents (ii) If circles touch internally `rArr C_(1)C_(2)=r_(1)+r_(2)`, 1 common tangent. (iii) If circles do not touch each other, 4 common tangents. Given equations of circles are `x^(2)+y^(2)-4x -6y-12=0" "...(i)` `x^(2)+y^(2)+6x+18y + 26=0" " ...(ii)` Centre of circle (i) is `C_(1)(2, 3)` and radius `=sqrt(4+9+12)=5(r_(1))" "["say"]` Centre of circle (ii) is `C_(2)(-3, -9)` and radius `=sqrt(9+81-26)=8(r_(2))" " ["say"]` Now, `C_(1)C_(2)=sqrt((2+3)^(2)+(3+9)^(2))` `rArr C_(1)C_(2)=sqrt(5^(2)+12^(2))` `rArrC_(1)C_(2)=sqrt(25+144)=13` `thereforer_(1)+r_(2)=5+8+13` Also `C_(1)C_(2)=r_(1)+r_(2)` Thus, both circles touch each other externally, Hence, there are three common tangents. |
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| 135. |
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` is |
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Answer» Correct Answer - `(3,-1)` Let point P be (h,k). Equation of chord of contact of the given circle w.r.t. point P is `4hx+4ky-3(x+h)+5(y+k)-14=0` or `(4h-3)x+(4k+5)y-3h+5k-14=0` (1) But the given equation of chord of contact is `9x+y-28=0` (2) `:. ` Comparing the ratio of coefficients of equation (1) and (2) we get `(4h-3)/(9) =(4k+5)/(1)=(3h-5k+14)/(28)` Solving we get `h=3` and `k= -1`. |
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| 136. |
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. `lamda(m_(2)+m_(3))+mu(m_(3)+m_(1))+v(m_(1)+m_(2))=0`B. `lamda(m_(2)m_(3)-1)+mu(m_(3)m_(1)-1)+v(m_(1)m_(2)-1)=0`C. Both (a) and (b)D. None of the above |
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Answer» Correct Answer - C |
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| 137. |
Find the equation of the circles which passes through the points ` A(1, 1) and B(2, 2 )` and whose radius is 1. Show that there are two such circles. |
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Answer» Let the centre of the circle be `C(h, k )`. Then, `|CA| = 1 and |CB| = 1 ` ` rArr ( h - 1 ) ^(2) + (k - 1 ) ^(2) = 1 ^(2) and ( h- 2 ) ^(2) + (k - 2 ) ^(2) = 1 ^(2)` `rArr h ^(2) + k ^(2) - 2h - 2k + 1 = 0 " "` ... (i) and ` h ^(2) + k ^(2) - 4h - 4k + 7 = 0 " " ` ... (ii) Subtacting (ii) from (i), we get ` 2h + 2k - 6 = 0 rArr h + k = 3 rArr k = ( 3- h ) ` Putting this value in (i), we get ` h ^(2) + ( 3- h ) ^(2) - 2h - 2 ( 3- h ) + 1 = 0 ` `rArr ( h - 2 ) (h - 1 ) = 0 rArr h = 2 or h = 1`. ` therefore k = 1 or k = 2 ` Hence, the centres are `(2, 1 ) and (1, 2)` |
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| 138. |
Find the other end of the diameter of the circle `x^(2) + y^(2) -8x-8y+27=0` if one end of it is `(2,3)`.A.B.C.D. |
| Answer» Correct Answer - The other end of the diameter is `B(6, 5) | |
| 139. |
Find the equation of the circle passing through (0, 0) and makingintercepts a and b on the coordinate axes. |
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Answer» Correct Answer - ` x ^(2) + y ^(2) - ax - by = 0 ` The circle passes through the points `O (0, 0), A(a, 0) and B(0, b)`. |
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| 140. |
Find the equation of the circle whose extremities of a diameter are `(1, 2)` and `(4, 5)`A.B.C.D. |
| Answer» Correct Answer - `x^(2) + y^(2) -5x -7y + 14 = 0` | |
| 141. |
Find the equation of the circle which passes through (1, 0) and (0, 1)and has its radius as small as possible. |
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Answer» The radius will be miniumum if the given points are the endpoints of a diameter. Then, the equation of the circle is `(x-1)(x-0)+(y-0)(y-1)=0` or ` x^(2)+y^(2)-x-y=0` |
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| 142. |
The equation of the circle which passes through `(1, 0)` and `(0, 1)` and has its radius as small as possible, 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` |
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Answer» Correct Answer - D |
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| 143. |
The circle `x^2 + y^2+ 4x-7y + 12 = 0` cuts an intercept on y-axis equal toA. 1B. 3C. 5D. 7 |
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Answer» Correct Answer - A |
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| 144. |
Find the equation of tangent to the circle x2 + y2 – 4x + 3y + 2 = 0 at the point (4, -2) |
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Answer» Given equation of the circle is x2 + y2 – 4x + 3y + 2 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = -4, 2f = 3, c = 2 g = -2, f = , c = 2 The equation of a tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 at (x1 , y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 The equation of the tangent at (4, -2) is x(4) + y(-2) – 2(x + 4) + 3/2 (y-2) + 2 = 0 ⇒ 4x – 2y – 2x – 8 + 3/2 y – 3 + 2 = 0 ⇒ 2x – 1/2y - 9 = 0 ⇒ 4x – y – 18 = 0 |
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| 145. |
`ABCD` is a square in first quadrant whose side is a, taking `AB and AD` as axes, prove that the equation to the circle circumscribing the square is `x^2+ y^2= a(x + y)`.A. `x^(2)+y^(2)+ax-ay=0`B. `x^(2)+y^(2)-ax+ay=0`C. `x^(2)+y^(2)-ax-ay=0`D. `x^(2)+y^(2)+ax-ay=0` |
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Answer» Correct Answer - C |
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| 146. |
Anequilateral triangle of side 9 cm is inscribed in a circle. Find the radiusof the circle. |
| Answer» let the equilateral triangle be ABC of each side of 9 cm prescribed in a circle.draw a perpendicular from A to BC , from C to AB, from B to AC respectively.now name the intersection point of perpendicular from A to BC as D,So, BD will be = `9/2` = 4.5let point of intersection of all perpendicular to centre of the circle as O.Now as we know that `/_OBD= 60^@` and `/_ODB=90^@`So, `/_DOB=180^@ - 60^@ - 90^@` = `30^@`now, `Cos 30^@` = `BD/BO``sqrt 3/ 2 = 4.5/BO`BO= `3*sqrt 3 ` | |
| 147. |
In the adjoining figure `angle QPO=24^(@)` and `angle SPR = 20^(@)` , find the value of `angle QOR`. |
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Answer» `angle QPR=25^(@)+20^(@)45^(@)` Now, the angle substended by an arc at centre is twice the angle subtended by the same arc at the remaining part of the circel. `therefore angle QOR=2angle QPR` `=2xx45^(@)` `90^(@)`. |
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| 148. |
The abscissa of the two points A and B are the roots of the equation `x^2+2a x-b^2=0`and their ordinates are the roots of the equation `x^2+2p x-q^2=0.`Find the equation of the circle with AB as diameter. Also, find itsradius. |
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Answer» Correct Answer - `x^(2)+y^(2)+2ax +2py-(b^(2)+q^(2)) = 0`. Let`(x_(1), y_(1)) and (x_(2), y_(2)) ` be the coordinates of points A and B, respectively. It is given that `x_(1)+x_(2) " are the roots of " x^(2)+2ax-b^(2)=0` `rArr" " x_(1)+x_(2)=-2a and x_(1)x_(2)=-b^(2)" "...(i)` Also, `y_(1) and y_(2) " are the roots of "y^(2)+2py-q^(2)=0` `rArr y_(1)+y_(2)=-2pand y_(1)y_(2)=-q^(2)" "...(ii)` `therefore` The equation of circle with AB as diameter is, `(x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0` `rArrx^(2)+y^(2)-(x_(1)+x_(2))x-(y_(1)+y_(2))y+(x_(1)x_(2)+y_(1)y_(2))=0` `rArr x^(2) + y^(2)+2ax+2py-(b^(2)+q^(2))=0` and radius `=sqrt(a^(2)+p^(2)+b^(2)+q^(2))` |
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| 149. |
If the abscissa and ordinates of two points `Pa n dQ`are the roots of the equations `x^2+2a x-b^2=0`and `x^2+2p x-q^2=0`, respectively, then find the equation of the circle with `P Q`as diameter. |
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Answer» Let `x_(1),x_(2)` and `y_(1),y_(2)` be the roots of `x^(2)+2ax-b^(2)=0` and `x^(2)+2px-q^(2)=0`, respectively. Then, `x_(1)+x_(2)=-2a,x_(1)x_(2)=-b^(2)` and `y_(1)+y_(2)=-2p,y_(1)y_(2)=-q^(2)` The equation of the circle with `P(x_(1),y_(1))` and `Q(x_(2),y_(2))` as the endpoints of diameter is `(x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0` or `x^(2)+y^(2)-x(x_(1)+x_(2))-y(y_(1)+y_(2))+x_(1)x_(2)+y_(1)y_(2)=0` or `x^(2)+y^(2)+2ax+2py-b^(2)-q^(2)=0` |
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| 150. |
Find the equation of the circle if the chord of the circle joining (1,2) and `(-3,1)`subtents `90^0`at the center of the circle. |
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Answer» The chord joining points A(1,2) and B(-3,1) subtends `90^(@)` at the center of the circle. Then AB subtend an angle `45^(@)` at point C on the circumference. Therefore, the equation of the circle is `(x-1)(x+3)+(y-2)(y-1)` `= +-cot 45^(@)[(y-2)(x+3)-(x-1)(y-1)]` or `x^(2)+y^(2)+2x-3y-1= +-[4y-x-7]` or `x^(2)+y^(2)+3x-7y+6=0` and `x^(2)+y^(2)+x+y-8=0` |
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