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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
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Verified Answer
The correct answer is:
$(-13,-4)$
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 $\mathrm{M}(\mathrm{x}, \mathrm{y})$
$$
\begin{aligned}
& \left(\frac{x+3}{2}, \frac{y-4}{2}\right)=(-5,-4) \\
& \Rightarrow M(-13,-4)
\end{aligned}
$$
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