JEE Advanced · Mathematics · 12. Circle

Let $T$ be the line passing through the points $P (- 2, 7)$ and $Q (2, - 5)$ . Let $F_{1}$ be the set of all pairs of circles $(S_{1}, S_{2})$ such that $T$ is tangents to $S_{1}$ at $P$ and tangent to $S_{2}$ at $Q$ , and also such that $S_{1}$ and $S_{2}$ touch each other at a point, say, $M$ . Let $E_{1}$ be the set representing the locus of $M$ as the pair $(S_{1}, S_{2})$ varies in $F_{1}$ . Let the set of all straight line segments joining a pair of distinct points of $E_{1}$ and passing through the point $R (1,1)$ be $F_{2}$ . Let $E_{2}$ be the set of the mid-points of the line segments in the set $F_{2}$ . Then, which of the following statement(s) is (are) TRUE?

A The point $(- 2,7)$ lies in $E_{1}$
B The point $(\frac{4}{5}, \frac{7}{5})$ does NOT lie in $E_{2}$
C The point $(\frac{1}{2}, 1)$ lies in $E_{2}$
D The point $(0, \frac{3}{2})$ does NOT lie in $E_{1}$

Verified Solution

Answer & Solution

Correct Answer

(D) The point $(0, \frac{3}{2})$ does NOT lie in $E_{1}$

Step-by-step Solution

Detailed explanation

A P = A Q = A M

∠ P M Q = 90 °

Locus of

M

is a circle having

P Q

as its diameter
Hence,

E_{1} : x^{2} + y^{2} - 2 y - 39 = 0, x \neq \pm 2

Clearly

(- 2, 7)

is not included in the locus because we can not consider point circles, and

(0, \frac{3}{2})

does not lie on

E_{1}

as it does not satisfy the equation above.

The equation of chord with

B (h, k)

as mid point will be given by

T = S_{1}

x h + y k - y - k - 39 = h^{2} + k^{2} - 2 k - 39

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match the following.

	List – I		List – II
(A)	Let $y (x) = \cos (3 \cos^{- 1} x), x ϵ [- 1, 1], x \neq \pm \frac{\sqrt{3}}{2} .$ Then $\frac{1}{y (x)} \{(x^{2} - 1) \frac{d^{2} y (x)}{d x^{2}} + x \frac{d y (x)}{d x}\}$ equals	(P)	1
(B)	Let $A_{1}, A_{2}, \dots, A_{n} (n > 2)$ be the vertices of a regular polygon of n sides with its centre at the origin. Let $\vec{a_{k}}$ be the position vector of the point $A_{k}, k = 1, 2, \dots n .$ If $\|\sum_{k = 1}^{n - 1} (\vec{a_{k}} \times \vec{a_{k + 1}})\| = \|\sum_{k = 1}^{n - 1} (\vec{a_{k}} \cdot \vec{a_{k + 1}})\|$ , then the minimum value of n is	(Q)	2
(C)	If the normal from the point $P (h, 1)$ on the ellipse $\frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$ is perpendicular to the line $x + y = 8,$ then the value of h is	(R)	8
(D)	Number of positive solutions satisfying the equation $\tan^{- 1} (\frac{1}{2 x + 1}) + \tan^{- 1} (\frac{1}{4 x + 1}) = \tan^{- 1} (\frac{2}{x^{2}})$ is	(S)	9

JEE Advanced 2014 Hard

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