SATHEE: CIRCLE-11 (Problem Solving)

CIRCLE-11 (Problem Solving)

Examples

1. Find the equation of the image of the circle $x^{2} + y^{2} + 16 x - 24 y + 183 = 0$ by the line mirror $4 x + 7 y + 13 = 0$

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Solution :

The given circle and line are

$x^{2} + y^{2} + 16 x - 24 y + 183 = 0$

(1) and $4 x + 7 y + 13 = 0$

Centre and radius if the circle are

$(- 8, 12)$ and $\sqrt{64 + 144 - 183} = \sqrt{25} = 5$ respectively.

Equation of line $C_{1} C_{2}$ is $7 x - 4 y + k = 0$ it passes through $(- 8, 12)$

$∴ - 56 - 48 + k = 0$

$k = 104$

Equation of line $C_{1} C_{2}$ is $7 x - 4 y + 104 = 0$

To get the coordinates of M. Solve the equation

(2) & (3)

(2) $\times 4 + (3) \times 7$

$16 x + 28 y + 52 = 0$
$49 x - 28 y + 728 = 0$
$65 x + 780 = 0$

$x = - 12$

put the value of $x$ in (2) we get

$- 48 + 7 y + 13 = 0$

$7 y = 35$

$y = 5$

$∴$ coordinate of $M$ is $(- 12, 5)$

$M$ is the midpoint of $C_{1}$ and $C_{2}$

$∴ - 12 = \frac{h - 8}{2} \Rightarrow h = - 16$

$5 = \frac{k + 12}{2} \Rightarrow K = - 2$

$∴$ Equation of imaged circle is

$(x + 16)^{2} + (y + 2)^{2} = 25$

$x^{2} + y^{2} + 32 x + 4 y + 235 = 0$

2. The circle passing through the point $(- 1, 0)$ and touching the $y$ -axis at $(0, 2)$ also passes through the point

(a) $(\frac{- 3}{2}, 10)$

(b) $(\frac{- 5}{2}, 12)$

(d) $(- 4, 0)$

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Solution:

Family of circles passing through a point $(0, 2)$ and touching line $x = 0$ (y-axis) is

$(x - 0)^{2} + (y - 2)^{2} + λ x = 0$

It passes through $(- 1, 0)$

$∴ 1 + 4 - λ = 0$

$∴ λ = 5$

$∴$ equation of circle is

$x^{2} + y^{2} + 5 x - 4 y + 4 = 0$

If also passes through $A (- x_{1}, 0)$

$∴ x_{1}^{2} - 5 x_{1} + 4 = 0$

$(x_{1} - 4) (x_{1} - 1) = 0$

$x_{1} = 4, x_{1} = 1$

$∴$ it also passes through $A (- 4, 0)$

3. Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3} + 1$ apart. If the chords subtend at the centre, angles of $\frac{π}{k}$ and $\frac{2 π}{k}$ , where $k > 0$ then the value of $[k]$ is…………..

( $[k]$ denotes the greatest integer)

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Solution:

Let $\frac{π}{2 k} = α = \frac{1}{2} ∠ A O B = ∠ A O M$

Then $∠ C O N = 2 α$

In $Δ A O M$

$\cos α = \frac{X}{2}$

In $Δ C O N$

$\cos 2 α = \frac{\sqrt{3} + 1 - x}{2}$

$\cos 2 α = 2 \cos^{2} θ - 1$

$\cos 2 α = 2 \frac{x^{2}}{4} - 1$

$∴ \frac{\sqrt{3} + 1 - x}{2} = \frac{x^{2}}{2} - 1$

$\begin{aligned} \sqrt{3} + 1 - x = x^{2} - 2 \\ x^{2} + x - \sqrt{3} - 3 = 0 \\ ∴ x = \frac{- 1 \pm \sqrt{1 + 4 (3 + \sqrt{3})}}{2} \\ = \frac{- 1 \pm \sqrt{13 + 4 \sqrt{3}}}{2} \\ = \frac{- 1 \pm (2 \sqrt{3} + 1)}{2} (∴ 13 + 4 \sqrt{3} = (2 \sqrt{3} + 1)^{2}) \\ x = \frac{- 1 + 2 \sqrt{3} + 1}{2} \\ x = \sqrt{3} \\ \cos α = \frac{\sqrt{3}}{2} = \cos \frac{π}{6} \\ α = \frac{π}{6} \\ Required angle = \frac{π}{k} = 2 α = \frac{π}{3} \Rightarrow k = 6 thus [k] = 6 \end{aligned}$

4. Let $A B C$ and ${A B}^{'}$ be two non-congruent triangles with sides $A B = 4, A C = {A C}^{'} = 2 \sqrt{2}$ and angle $β = 30^{\circ}$ . The absolute value of the difference between the areas of these triangles is

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Solution:

Draw circle through ${A C}^{'}$ and $A B$ intersect the circle and $P$

$\begin{aligned} {I n}_{Δ} A B D \\ = \frac{A D}{A B} = \sin 30^{\circ} \\ \frac{A D}{4} = \frac{1}{2} \\ ∴ A D = 2 = D C = C^{'} D \end{aligned}$

Difference of areas of $△ A B C$ and $_{Δ} A B C^{'}$ is $Δ {A C C}^{'}$

$∴ ar (Δ {A C C}^{'}) = \frac{1}{2} \times 4 \times 2 = 4$ sq.u.

5. The centres of two circles $C_{1}$ and $C_{2}$ each of unit radius are at a distance of 6 units from each other. Let $P$ be the mid-point of the line segment joining the centres of $C_{1}$ and $C_{2}$ and $C$ be a circle touching circles $C_{1}$ and $C_{2}$ externally. If a common tangents to $C_{1}$ and $C$ passing through $P$ is also a common tangent to $C_{2}$ and $C_{1}$ then the radius of the circle $C$ is

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Solution :

In $△ {C P C}_{2}$

${C P}^{2} = {({C C}_{2})}^{2} - {(C_{2} P)}^{2}$

$k^{2} = (r + 1)^{2} - 9$

$k^{2} = r^{2} + 2 r - 8 \dots \dots . (1)$

In $Δ {P Q C}_{2}$

$P Q^{2} = 3^{2} - 1^{2}$

$= 8$

$∴$ In $_{Δ} C P Q$

$k^{2} = r^{2} + 8 \dots \dots \dots . . (2)$

From (1) & (2)

$r^{2} + 8 = r^{2} + 2 r + 2 r - 8. .$

$16 = r$

$∴ r = 8$

6. A straight line through the vertex $P$ of a triangle $P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of the triangle $P Q R$ at the point $T$ . If $S$ is not the centre of the circumcircle then

(a) $\frac{1}{P S} + \frac{1}{S T} < \frac{2}{\sqrt{Q S \cdot S R}}$

(b) $\frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S . S R}}$

(d) $\frac{1}{P S} + \frac{1}{S T} > \frac{4}{Q R}$

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Solution :

Points P, Q, T, R are concyclic

$∴$ PS.ST $=$ QS.SR

$\Rightarrow \frac{P S + S T}{2} \geq \sqrt{P S . S T} (A M \geq G M)$

$∴ P T \geq 2 \sqrt{P S . S T}$

and $\frac{1}{P S} + \frac{1}{S T} \geq \frac{2}{\sqrt{P S . S T}} = \frac{2}{\sqrt{Q S . S R}}$

Also, $\frac{S Q + S R}{2} \geq \sqrt{SQ.SR}$

$\Rightarrow \frac{Q R}{2} \geq \sqrt{S Q . S R}$

$\Rightarrow \frac{1}{\sqrt{SQ.SR}} \geq \frac{2}{Q R}$

$\Rightarrow \frac{2}{\sqrt{SQ.SR}} \geq \frac{4}{Q R}$ $(A M \geq G M)$

(Dividing by PS.ST)

$\frac{1}{P S} + \frac{1}{S T} \geq \frac{2}{\sqrt{Q S . S R}} \geq \frac{4}{Q R}$

7. Let $A B C D$ be a quadrilateral with area 18 , with side $A B$ parallel to the side $C D$ and $A B = 2 C D$ . Let $A D$ be perpendicular to $A B$ and $C D$ . If a circle is drawn inside the quadrilateral $A B C D$ touching all the sides then its radius is

(a) $3$

(b) $2$

(d) $1$

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Solution :

$A B C D$ is a trapezium ( $A B$ parallel to $C D$ )

$∴ ar (A B C D) = \frac{1}{2} \times h$ (sum of parallel sides)

$= \frac{1}{2} \times 2 r (2 a + a)$

$18 = r \times 3 a$

ar $= 6$

$C B$ is a tangent to the circle

$∴$ equation of tangent is

$y = \frac{- 2 r}{a} (x - 2 a) \Rightarrow 2 r x + a y - 4 a r = 0$

It is a tangent to the circle $(x - r)^{2} + (y - r)^{2} = r^{2}$

$∴ r = | \frac{2 r^{2} + a r - 4 a r}{\sqrt{4 r^{2} + a^{2}}} |$

$r \sqrt{4 r^{2} + a^{2}} = 2 r^{2} - 3 a r$

$\sqrt{4 r^{2} + a^{2}} = 2 r - 3 a$

Squaring

$4 r^{2} + a^{2} = 4 r^{2} + 9 a^{2} - 12 a r$

$12 r = 8 a \Rightarrow 3 r = 2 a$

ar $= 6$

$r = \frac{29}{3}$

$\frac{2 a^{2}}{3} = 6$

$a^{2} = 9$

$a = \pm 3$

$∴ r = 2$

8. The radius of the least circle passing through the point $(8, 4)$ and cutting the circle $x^{2} + y^{2} = 40$ orthogonally is

(a) $\sqrt{5}$

(b) $\sqrt{7}$

(d) $3 \sqrt{5}$

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Solution :

Let the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

Given circle is $x^{2} + y^{2} = 40$

These two circles are orthogonal

$∴ c - 40 = 0 \Rightarrow c = 40$

(1) passes through $(8, 4)$

$64 + 16 + 16 g + 8 f + 40 = 0$

$120 + 16 g + 8 f = 0$

$f + 2 g + 15 = 0$ or $f = - (2 g + 15)$

radius $= \sqrt{g^{2} + f^{2} - c}$

$= \sqrt{g^{2} + (2 g + 15)^{2} - 40}$

For least circle radius must be minimum

Let $f (g) = g^{2} + (2 g + 15)^{2} - 40$ is minimum

$f^{'} (g) = 2 g + 4 (2 g + 15) = 0$

$10 g = - 60$

$g = - 6$

$f^{''} (g) = 10 > 0$ minimum

$f = - (- 12 + 15) = - 3$

Equation of circle in $x^{2} + y^{2} - 12 x - 6 y + 40 = 0$

radius $= \sqrt{36 + 9 - 40}$

$= \sqrt{5}$

9. $P$ is a point (a,b) in the first quadrant. If the two circles which pass throngh $P$ and touch both the co-ordinate axes cut at right angles, then

(a) $a^{2} - 6 a b + b^{2} = 0$

(b) $a^{2} + 2 a b - b^{2} = 0$

(d) $a^{2} - 8 a b + b^{2} = 0$

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Solution :

Equation of the two circles be

$(x - r)^{2} + (y - r)^{2} = r^{2} \Rightarrow x^{2} + y^{2} - 2 x r - 2 y r + r^{2} = 0$

These two circles passes through $(a, b)$

$∴ (a - r)^{2} + (b - r)^{2} = r^{2}$

$a^{2} + r^{2} - 2 a r + b^{2} + r^{2} - 2 b r - r^{2} = 0$

$r^{2} - 2 r (a + b) + (a^{2} + b^{2}) = 0$

It is a quadratic equation in $r$

$∴ r_{1} + r_{2} = 2 (a + b)$ and $r_{1} r_{2} = a^{2} + b^{2}$

Condition for orthogonality is

$2 g_{1} g_{2} + 2 f_{1} f_{2} = C_{1} + C_{2}$

$2 r_{1} r_{2} + 2 r_{1} r_{2} = r_{1}^{2} + r_{2}^{2}$

$4 r_{1} r_{2} = r_{1}^{2} + r_{2}^{2}$

$6 r_{1} r_{2} = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2}$

$6 r_{1} r_{2} = {(r_{1} + r_{2})}^{2}$

$6 (a^{2} + b^{2}) = 4 (a + b)^{2}$

$\begin{aligned} 6 a^{2} + 6 b^{2} = 4 a^{2} + 4 b^{2} + 8 a b \\ 2 a^{2} + 2 b^{2} - 8 a b = 0 \\ a^{2} + b^{2} - 4 a b = 0 \end{aligned}$

10. A circle $S_{\equiv} 0$ passes throngh the common points of family of circles $x^{2} + y^{2} + d x - 4 y + 3 = 0 (λ \in R)$ and have minimum area then

(a) area of $S \equiv 0$ is $π$ squ.

(b) radius of director circle of $S \equiv 0$ is $\sqrt{2}$

(d) $S \equiv 0$ never cuts $| 2 x | = 1$

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Solution :

$(x^{2} + y^{2} - 4 y + 3) + λ x = 0$

$∴ x = 0$ and $y^{2} - 4 y + 3 = 0$

$(y - 3) (y - 1) = 0$

$y = 3, 1$

$∴ (0, 3) (0, 1)$ are common points.

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

passion through $(0, 3) & (0, 1)$

$9 + 6 f + c = 0$

$1 + 2 f + c = 0$

(1) - (2) we get

$8 + 4 f = 0$

$f = - 2$ and $c = 3$

$∴ x^{2} + y^{2} + 2 g x - 4 y + 3 = 0$

radius $= \sqrt{g^{2} + 4 - 3} = \sqrt{g^{2} + 1}$

for minimum area radius must be minimum

Since $g^{2} + 1$ is positive so $g$ must be zero

$∴$ radius $= 1$

Area $= π r^{2} = π$ sq.u.

Radius of director circle is $\sqrt{2}$ times the radius of the given circle.

$∴$ Radius of director circle is $\sqrt{2}$

11. Area of part of circle $x^{2} + y^{2} - 4 x - 6 y + 12 = 0$ above the line $4 x + 7 y - 29 = 0$ is $Δ$ , then $[Δ] =$ [.] is greatest integer function.

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Solution :

Since line $4 x + 7 y - 29 = 0$

passes through the centre $(2, 3)$ of the circle

$∴$ The line is a diameter of a circle with radius $\sqrt{4 + 9 - 12} = 1$

$∴$ area of semi circle is $= \frac{1}{2} π r^{2}$

$= \frac{1}{2} π$

$= \frac{3.14}{2} = 1.57$

Hence $[Δ] = [1.57] = 1$

Practice questions

1. Let $0 < α < \frac{π}{2}$ be a fixed angle. If $P = (\cos θ, \sin θ)$ and $Q = (\cos (α - θ), \sin (α - θ) Q$ is obtained from $P$ by

(a). clockwise rotation around origin through an angle $α$

(b). anti-clockwise rotation around origin through an angle $α$

(c). reflection in the line through origin with slope $\tan α$

(d). reflection in the line through origin with slope tan $α / 2$

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Answer: (d)

2. If the tangent at the point $P$ on the circle $x^{2} + y^{2} + 6 x + 6 y = 2$ meets the straight line $5 x - 2 y + 6 = 0$ at a point $Q$ on the $y$ -axis, then the length of $P Q$ is

(a). 4

(b).

(c). 5

(d). $3 \sqrt{5}$

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Answer: (c)

3. The equations to the sides $A B, B C, C A$ of a $△ A B C$ are drawn on $A B, B C, C A$ as diameters. The point of concurrence of the common chord is

(a). centroid of the triangle

(b). orthocenter

(c). circumcentre

(d). incentre

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Answer: (b)

4. The number of rational point(s) (a point $(a, b)$ is rational, if $a$ and $b$ both are rational numbers) on the circumference of a circle having centre $(π, e)$ is

(a). at most one

(b). at least two

(c). exactly two

(d). infinite

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Answer: (a)

5. The locus of a point such that the tangents drawn from it to the circle $x^{2} + y^{2} - 6 x - 8 y = 0$ are perpendicular to each other is

(a). $x^{2} + y^{2} - 6 x - 8 y - 25 = 0$

(b). $x^{2} + y^{2} + 6 x - 8 y - 5 = 0$

(c). $x^{2} + y^{2} - 6 x + 8 y - 5 = 0$

(d). $x^{2} + y^{2} - 6 x - 8 y + 25 = 0$

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Answer: (a)

6. If the two circles $x^{2} + y^{2} + 2 g x + 2 f y = 0$ and $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0$ touch each other, then

(a). $f_{1} g = f g_{1}$

(b). ${f f}_{1} = {g g}_{1}$

(c). $f^{2} + g^{2} = f_{1}^{2} + g_{1}^{2}$

(d). none of these

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Answer: (a)

7. The number of integral values of $λ$ for which $x^{2} + y^{2} + λ x + (1 - λ) y + 5 = 0$ is the equation of a circle whose radius cannot exceed 5 , is

(a). 14

(b). 18

(c). 16

(d). none of these

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Answer: (c)

8. The circle $x^{2} + y^{2} + 4 x - 7 y + 12 = 0$ cuts an intercept on $y$ -axis of length

(a). 3

(b). 4

(c). 7

(d). 1

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Answer: (d)

9. One of the diameter of the circle $x^{2} + y^{2} - 12 x + 4 y + 6 = 0$ is given by

(a). $x + y = 0$

(b). $x + 3 y = 0$

(c). $x = y$

(d). $3 x + 2 y = 0$

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Answer: (b)

10. The coordinates of the middle point of the chord cut off by $2 x - 5 y + 18 = 0$ by the circle $x^{2} + y^{2} - x + y 2 - 54 = 0$ are

(a). $(1, 4)$

(b). $(2, 4)$

(c). $(4, 1)$

(d). $(1, 1)$

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Answer: (a)

11. A variable chord is drawn through the origin to the circle $x^{2} + y^{2} - 2 a x = 0$ . The locus of the centre of the circle drawn on this chord as diameter is

(a). $x^{2} + y^{2} + a x = 0$

(b). $x^{2} + y^{2} + a y = 0$

(c). $x^{2} + y^{2} - a x = 0$

(d). $x^{2} + y^{2} - a y = 0$

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Answer: (c)

12. If $O$ is the origin and $O P, O Q$ are distinct tangents to the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , the circumcentre of the triangle $O P Q$ is

(a). $(- g, - f)$

(b). $(g, f)$

(c). $(- f, - g)$

(d). none of these

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Answer: (d)

13. Equation of the normal to the circle $x^{2} + y^{2} - 4 x + 4 y - 17 = 0$ which passes through $(1, 1)$ is

(a). $3 x + 2 y - 5 = 0$

(b). $3 x + y - 4 = 0$

(c). $3 x + 2 y - 2 = 0$

(d). $3 x - y - 8 = 0$

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Answer: (b)

14. The equation of the circle touching the lines $| y | = x$ at a distance $\sqrt{2}$ unit from the origin is

(a). $x^{2} + y^{2} - 4 x + 2 = 0$

(b). $x^{2} + y^{2} + 4 x - 2 = 0$

(c). $x^{2} + y^{2} + 4 x + 2 = 0$

(d). none of these

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Answer: (a)

15. The shortest distance from the point $(2, - 7)$ to the circle $x^{2} + y^{2} - 14 x - 10 y - 151 = 0$ is

(a). 1

(b). 2

(c). 3

(d). 4

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Answer: (b)

16. The equation of the image of the circle $(x - 3)^{2} + (y - 2)^{2} = 1$ by the mirror $x + y = 19$ is

(a). $(x - 14)^{2} + (y - 13)^{2} = 1$

(b). $(x - 15)^{2} + (y - 14)^{2} = 1$

(c). $(x - 16)^{2} + (y - 15)^{2} = 1$

(d). $(x - 17)^{2} + (y - 16)^{2} = 1$

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Answer: (d)

17. If $P$ and $Q$ are two points on the circle $x^{2} + y^{2} - 4 x - 4 y - 1 = 0$ which are farthest and nearest respectively from the point $(6, 5)$ , then

(a). $P = - \frac{22}{5}, 3$

(b). $Q = \frac{22}{5}, \frac{19}{5}$

(c). $P = \frac{14}{3}, - \frac{11}{5}$

(d). $Q = - \frac{14}{3}, - 4$

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Answer: (b)

18. A circle of the coaxial system with limiting points $(0, 0)$ and $(1, 0)$ is

(a). $x^{2} + y^{2} - 2 x = 0$

(b). $x^{2} + y^{2} - 6 x + 3 = 0$

(c). $x^{2} + y^{2} = 1$

(d). $x^{2} + y^{2} - 2 x + 1 = 0$

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Answer: (d)

19. If a variable circle touches externally two given circles, then the locus of the centre of the variable circle is

(a). a straight line

(b). a parabola

(c). an ellipse

(d). a hyperbola

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Answer: (d)

Passage - 1

Let $A \equiv (a, 0)$ and $B$ be two fixed points and $P$ moves on a plane such that $P A = n P B$

On the basis of above information, answer the following questions:

20. If $| n | \neq 1$ , then the locus of a point $P$ is

(a). a straight line

(b). a circle

(c). a parabola

(d). an ellipse

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Answer: (b)

21. If $n = 1$ , then the locus of a point $P$ is

(a). a straight line

(c). a circle

(c). a parabola

(d). a hyperbola

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Answer: (a)

22. If $0 < n < 1$ , then

(a). A lies inside the circle and $B$ lies outside the circle

(b). A lies outside the circle and $B$ lies inside the circle

(c). both $A$ and $B$ lies on the circle

(d). both $A$ and $B$ lies inside the circle

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Answer: (a)

23. If $n > 1$ , then

(a). A lies outside the circle and $B$ leis inside the circle

(b). A lies outside the circle and $B$ leis inside the circle

(c). both $A$ and $B$ lies on the circle

(d). both $A$ and $B$ lies inside the circle

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Answer: (b)

24. If focus of $P$ is a circle, then the circle

(a). passes through $A$ and $B$

(b). never passes through $A$ and $B$

(c). passes through $A$ but does not pass through $B$

(d). passes through B but does not pass through A

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Answer: (b)

Passage - 2

For each natural number $k$ , let $C k$ denotes the circle with radius $k$ units and centre at the origin. On the $C k e$ , a particle moves $k$ units in the counter clockwise direction. After completing its motion on $C k$ , the particle moves to $C_{k t 1}$ in some well defined manner, where $k > 0$ . The motion of the particle continues in this manner.

On the basis of above information, answer the following questions:

25. Let, $K = 1$ the particle starts at $(1, 0)$ . If the particle crossing the positive direction of the $x$ -axis for the first time on the circle $C n$ , then $n$ is equal to

(a). 3

(b). 5

(c). 7

(d). 8

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Answer: (c)

26. If $k \in N$ and , the particle starts $(- 1, 0)$ the particle cross $x$ -axis again at

(a). $(3, 0)$

(b). $(1, 0)$

(c). $(4, 0)$

(d). $(2, 0)$

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Answer: (c)

27. If and , the particle moves in the radial direction from circle $C k$ to $C K + 1$ . If particle starts form the point $(- 1, 0)$ , then

(a). it will cross the + ve $y$ -axis at $(0, 4)$

(b). it will cross the - ve $y$ -axis at $(0, - 4)$

(c). it will cross the + ve $y$ -axis at $(0, 5)$

(d). it will cross the - ve $y$ -axis at $(0, - 5)$

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Answer: (c)

28. If and , particle moves tangentially form the circle $C k$ to $C k + 1$ , such that the length of tangent is equal to $k$ units itself. If particle starts form the point $(1, 0)$ , then

(a). the particle will cross $x$ -axis again at $x = 3$

(b). the particle will cross $x$ -axis again at $x = 4$

(c). the particle will cross +ve $x$ -axis again at $x =$

(d). the particle will cross +ve $x$ -axis again at $x \in (2 \sqrt{2}, 4)$

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Answer: (d)

29. Let the particle starts from the point $(2, 0)$ and moves $π / 2$ units, on circle $C 2$ in the counterclockwise direction, then moves on the circle $C 3$ along the tangential path, let this straight line (tangential path traced by particle) intersect the circle $C 3$ at the points $A$ and $B$ tangents drawn at $A$ and $B$ intersect at

(a). $\frac{9}{2 \sqrt{2}}; \frac{9}{2 \sqrt{2}}$

(b). $(9 \sqrt{2}, 9 \sqrt{2})$

(c). $(9, 9)$

(d). $(\sqrt{2}, \sqrt{2})$

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Answer: (a)

Match Type:

30. Observe the following columns:

Columns I	Columns II
(a). If the shortest and largest distance from the point $(10, 7)$ to the circle $x^{2} + y^{2} - 4 x - 2 y - 20 = 0$ are $L$ and $M$ respectively, then	p. $M + L = 10$
(b). If the shortest and largest distance from the point $(3, - 6)$ to the circle $x^{2} + y^{2} - 16 x - 12 y - 125 = 0$ are $L$ and $M$ respectively, then	q. $M + L = 20$
(c). If the shortest and largest distance from the point $(6, - 6)$ to the circle $x^{2} + y^{2} - 4 x + 6 y - 20 = 0$ are $L$ and $M$ respectively, then	r. $M + L = 30$
	s. $M - L = 10$
	t. $M - L = 26$

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Answer: a $\rarr$ q, s; b $\rarr$ r, t; c $\rarr$ s

31. Observe the following columns:

Columns I	Columns II
(a). If the straight lines $y = a_{1} x + b$ and $y = a_{2} x + b (a_{1} \neq a_{2})$ meet the coordinate axes in concyclic points, then	p. $a_{1}^{2} + a_{2}^{2} = 4$
(b). If the chord of contact of the tangents drawn to $x^{2} + y^{2} = b^{2}$ from any point on $x 2 + y 2 =$ , touches the circle $x 2 + y 2 = a_{2}^{2} (a_{1} \neq a_{2})$ , then	q. $a 1 + a 2 = 3$
(c). If the circles $x^{2} + y^{2} + 2 a_{1} x + b = 0$ and $x^{2} + y^{2} + 2 a_{2} x + b = 0 (a_{1} \neq a_{2})$ and orthogonally, then	r. $a_{1} a_{2} = b$
	s. $a_{1} a_{2} = 1$
	t. $a_{1} a_{2} = b^{2}$

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Answer: a $\rarr$ p, q, s; b $\rarr$ p, q, s, t; c $\rarr$ p, q, r, s

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CIRCLE-11 (Problem Solving)

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CIRCLE-11 (Problem Solving)

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