Which of the following is the only CORRECT combination? [2019]

Question

Let the circles $C_{1} : x^{2} + y^{2} = 9$ and $C_{2} : {(x - 3)}^{2} + {(y - 4)}^{2} = 16$ , intersect at the points $X$ and $Y$ . Suppose that another circle $C_{3} : {(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ satisfies the following conditions:

(i) Centre of $C_{3}$ is collinear with the centres of $C_{1}$ and $C_{2}$ ,

(ii) $C_{1}$ and $C_{2}$ both lie inside $C_{3}$ , and

(iii) $C_{3}$ touches $C_{1}$ at $M$ and $C_{2}$ at $N$ .

Let the line through $X$ and $Y$ intersect $C_{3}$ at $Z$ and $W$ , and let a common tangent of $C_{1}$ and $C_{3}$ be a tangent to the parabola $x^{2} = 8 α y$ .

There are some expressions given in the Column-I whose values are given in Column-II below:

	Column I		Column II
(A)	$2 h + k$	(p)	$6$
(B)	$\frac{Length of Z W}{Length of X Y}$	(q)	$\sqrt{6}$
(C)	$\frac{Area of triangle M Z N}{Area of triangle Z M W}$	(r)	$\frac{5}{4}$
(D)	$α$	(s)	$\frac{21}{5}$
		(t)	$2 \sqrt{6}$
		(u)	$\frac{10}{3}$

Q. Which of the following is the only CORRECT combination? [2019]

Smart Achievers · Accepted Answer

(4)

$Given three circles are$

$C_{1} : x^{2} + y^{2} = 9$

$C_{2} : {(x - 3)}^{2} + {(y - 4)}^{2} = 16$

$C_{3} : {(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

$Centres of circles C_{1}, C_{2}, C_{3} are D (0, 0), E (3, 4), F (h, k) respectively$

$and radii of circles C_{1} : C_{2} : C_{3} are 3, 4, r respectively$

$Equation of D E : y = \frac{4}{3} x$

$Centres of circles C_{1}, C_{2}, C_{3} are collinear \Rightarrow F (h, \frac{4}{3} h)$

$M N = M D + D E + E N = 3 + 5 + 4 = 12 \Rightarrow r = 6$

$∴$ $D E = 6 - 3 = 3$

$\Rightarrow h^{2} + \frac{16}{9} h^{2} = 9 \Rightarrow h^{2} = \frac{81}{25} \Rightarrow h = \frac{9}{5}$ $taking h +ve, as lies between D and E$

$∴$ $F (\frac{9}{5}, \frac{12}{5})$

$∴$ $2 h + k = \frac{18}{5} + \frac{12}{5} = \frac{30}{5} = 6$

$DE is common chord of circles C_{1} and C_{2}$

$∴$ $Equation of X Y : S_{1} - S_{2} = 0$

$\Rightarrow 6 x + 8 y - 18 = 0 \Rightarrow 3 x + 4 y - 9 = 0$

$Length of perpendicular from D to X Y = \frac{9}{5} = D P$

$Also D X = 3, P X = \sqrt{9 - \frac{81}{25}} = \sqrt{\frac{225 - 81}{25}} = \frac{12}{5}$

$∴$ $X Y = 2 P X = \frac{24}{5}$

$ZW is chord of C_{3}$

$F P = M F - M P = 6 - (3 + \frac{9}{5}) = 6 - \frac{24}{5} = \frac{6}{5}$

$∴$ $Z P = \sqrt{6^{2} - {(\frac{6}{5})}^{2}} = \frac{12 \sqrt{6}}{5}$ $∴$ $Z W = \frac{24 \sqrt{6}}{5}$

$Hence, \frac{Length of Z W}{Length of X Y} = \frac{24 \sqrt{6} / 5}{24 / 5} = \sqrt{6}$

$∴$ $(B) - (q)$

$Area of ∆ M Z N = \frac{1}{2} \times M N \times Z P = \frac{1}{2} \times 12 \times \frac{12 \sqrt{6}}{5} = \frac{72 \sqrt{6}}{5}$

$Area of ∆ Z M W = \frac{1}{2} \times Z W \times M P = \frac{1}{2} \times \frac{24 \sqrt{6}}{5} \times \frac{24}{5} = \frac{288 \sqrt{6}}{25}$

$∴$ $\frac{Area of ∆ M Z N}{Area of ∆ Z M W} = \frac{72 \sqrt{6}}{5} \times \frac{25}{288 \sqrt{6}} = \frac{5}{4}$

$∴$ $(C) - (r)$

$Now common tangent of C_{1} and C_{3} is S_{1} - S_{3} = 0$

$\Rightarrow 2 h x + 2 k y - h^{2} - k^{2} = 9 - r^{2}$

$\Rightarrow \frac{18}{5} x + \frac{24}{5} y - \frac{81}{25} - \frac{144}{25} = 9 - 36$

$\Rightarrow 3 x + 4 y + 15 = 0$

$It is tangent to x^{2} = 8 α y$

Putting value of $y$ from common tangent in parabola, we get

$\Rightarrow x^{2} = - 8 α (\frac{3 x + 15}{4})$ $\Rightarrow x^{2} + 6 α x + 30 α = 0$

It should have equal roots

$∴$ $36 α^{2} - 120 α = 0 \Rightarrow α = \frac{10}{3}$

$∴$ $(D) - (u)$

Thus $(B) - (q)$ is the only correct combination and $(D) - (s)$ is the only incorrect combination.

Option (4) is correct

Solution

1 $(A), (u)$

2 $(A), (s)$

3 $(B), (t)$

4 $(B), (q)$

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Solution

1 (A),(u)

2 (A),(s)

3 (B),(t)

4 (B),(q)

Want Us to Email you About Special Offers & Updates?

1 $(A), (u)$

2 $(A), (s)$

3 $(B), (t)$

4 $(B), (q)$