Two sides of a square are along the lines $x=-5$ and $y=4$. The point of intersection of the diagonals is $(3,-4)$. The point of intersection of the tangents drawn to the circumcircle of the square at the two consecutive vertices lying on $x=-5$ is

• A

  $(-4,-4)$
• Correct answer
  B

  $(-13,-4)$
• C

  $(-4,-13)$
• D

  $(-4,-10)$

Solution

For point $\mathrm{R}(\mathrm{x}, \mathrm{y})$ $$ \begin{aligned} & \left(\frac{\mathrm{x}-5}{2}, \frac{\mathrm{y}+4}{2}\right)=(3,-4) \\ & \Rightarrow \mathrm{R}(11,-12) \end{aligned} $$ Since PQRS is square $\therefore \mathrm{Q}(-5,-12)$ and $\mathrm{S}(4,11)$ Also PMQC is square $$ \therefore \mathrm{N}(\mathrm{x}, \mathrm{y})=\left(\frac{-5-5}{2}, \frac{4,-12}{2}\right)=(-5,-4) $$ For point $\mathrm{M}(\mathrm{x}, \mathrm{y})$ $$ \begin{aligned} & \left(\frac{x+3}{2}, \frac{y-4}{2}\right)=(-5,-4) \\ & \Rightarrow M(-13,-4) \end{aligned} $$

Hence, the correct answer is option: 'B'.