Consider three circles $C_{1}, C_{2}$ and $C_{3}$ such that $C_{2}$ is director circle of $C_{1}$ and $C_{3}$ director circle of $C_{2}$ . 2 Tangents are drawn to $C_{1}$ from a point ‘A’ on $C_{3}$ which intersect $C_{2}$ at P and Q (take P and Q which are nearer to A), then which of the following is/are correct?

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The angle between the tangents to $C_{2}$ at P and Q is $\frac{5 π}{6}$

The angle between the tangents to $C_{2}$ at P and Q is $\frac{π}{6}$

Locus of point of intersection of tangents at P and Q is circle

Locus of point of intersection of tangents at P and Q is ellipse

answer is A, C.

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Detailed Solution

Let $C_{1}$ be a circle of radius r.
So, the radius of $C_{2}$ will be $\sqrt{2} r$ and that of $C_{3}$ will be 2r.
In $Δ A O T$ ,
$\sin \frac{θ}{2} = \frac{r}{2 r} = \frac{1}{2}$ or $θ = \frac{π}{3}$
Or $∠ A T B = \frac{π}{3}$
In $Δ A O P$ ,
$\sin ∠ A P O = \frac{r}{r \sqrt{2}} o r ∠ A P O = \frac{π}{4}$
In $Δ O P H, ∠ P O T = \frac{π}{4} - \frac{π}{6} = \frac{π}{12}$
In $Δ O P H$ ,
$Δ O P H = \frac{π}{2}$
$∠ P H O = π - (\frac{π}{2} + \frac{π}{12}) = \frac{5 π}{12}$
Or $∠ P H Q = \frac{5 π}{6}$
As $∠ P O Q = \frac{π}{6}, O P H Q$ is a cyclic quadrilateral.
Locus of H is a circle.