If the area of the triangle formed by the positive $x$-axis, the normal and the tangent to the circle $(x-2)^{2}+(y-3)^{2}=25$ at the point $(5,7)$ is $\mathrm{A}$, then $24 \mathrm{~A}$ is equal to
Note: NTA has dropped this question in the final official answer key.
Equation of normal at $\mathrm{P}$
$(y-7)=\left(\frac{7-3}{5-2}\right)(x-5)$
$3 y-21=4 x-20$
$\Rightarrow 4 x-3 y+1=0 \ldots(1)$
$\Rightarrow M\left(-\frac{1}{4}, 0\right)$
Equation of tangent at $\mathrm{P}$
$(y-7)=-\frac{3}{4}(x-5)$
$4 y-28=-3 x+15$
$\Rightarrow 3 x+4 y=43 \ldots \ldots$ (ii)
$\Rightarrow \quad \mathrm{N}\left(\frac{43}{3}, 0\right)$
Hence ar $(\Delta$ PMN $)=\frac{1}{2} \times$ MN $\times 7$
$\lambda=\frac{1}{2} \times \frac{175}{12} \times 7$
$\rightarrow 24 \lambda=1225$
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