Question:
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 $A$, then $24 \mathrm{~A}$ is equal to
Solution:
Equation of normal
$4 x-3 y+1=0$
and equation of tangents
$3 x+4 y-43=0$
Area of triangle $=\frac{1}{2}\left(\frac{43}{3}+\frac{1}{4}\right) \times(7)$
$=\frac{1}{2}\left(\frac{172+3}{12}\right) \times 7$
$A=\frac{1225}{24}$
$24 \mathrm{~A}=1225$
* as positive $x$-axis is given in the question so question should be bonus.