Question

Two tangents are drawn on the circle $x^2+y^2-6x-2y-15=0$ at point $B(3,6)$ and $D(0,-3)$ which meets at point $C$. If $A$ be the center of circle, then area of quadrilateral $ABCD$ is

Answer 1

Equation of the tangent to the circle
$x^2+y^2+2gx+2fy+c=0$ at $(x_1,y_1)$

$x{x}_1+y{y}_1+g(x+x_1)+f(y+y_1)+c=0$
Equation of $BC$ at $B(3,6)$
$y=6$
Equation of $CD$ at $D(0,-3)$
$3x+4y+12=0$
So, the intersection point of $BC$ and $CD$ is,
$C=(-12,6)$
$BC=15$ and $AB=5$
Area of quadrilateral $ABCD$ $=\frac{1}{2}\times 15\times 5\times 2=75$