Common tangents are drawn to two circles $x^{2} + y^{2} = 1$ and ${(x - 4)}^{2} + {(y - 4)}^{2} = 4.$ If the intersection points of transverse common tangents with the direct common tangents are $P, Q, R, S$ then which is/are correct

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equation of circumcircle of $Δ P Q R$ is $x^{2} + y^{2} - 4 x - 4 y = 0$

$P Q R S$ is a cyclic quadrilateral

The circumcircle of $Δ Q R S$ touches x-axis

$P Q R S$ is a square

answer is B, C, D.

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Detailed Solution

$\begin{array}{l} ∠ C_{1} P C_{2} = ∠ C_{1} Q C_{2} = ∠ C_{1} R C_{2} = ∠ C_{1} S C_{2} = 90 ° \\ ∴ P, Q, R, S a r e c o n c y l i c & c i r c l e t h r o u g h t h e m i s \bar{C_{1} C_{2}} \\ d i a m e t e r c i r c l e w h o s e e q u a t i o n i s x^{2} + y^{2} - 4 x - 4 y = 0 \end{array}$