Question

Let $A$ be the centre of the circle $x^2+y^2-2x-4y-20=0$. The tangents at the points $B(1,7)$ and $D(4,-2)$ on the circle meet at the point $C$. If $△$ is the area of the quadrilateral $ABCD$, then $\frac{△}{15}$ is equal to

Answer 1

The given equation of circle is
$x^2+y^2-2x-4y-20=0$
Here, $g=-1,f=2$
Center is $(1,2)$
So, the coordinates of point $A$ are $(1,2)$.
Now, the tangent to the circle at $B(1,7)$ is
$x+7y-(x+1)-2(y+7)-20=0$
$\Rightarrow y=5$
Tangent to circle at $C(4,-2)$ is
$4x-2y-(x+4)-2(y-2)-20=0$
$\Rightarrow 3x-4y-20=0$
These tangents intersect at $D(16,7)$.
Area of the quadrilateral $ABCD=2$ Area of the triangle $ABD$
Area of quadrilateral $=\Delta =\left|\begin{array}{ccc}1& 2& 1\\ 1& 7& 1\\ 16& 7& 1\end{array}\right|=75$ sq.units
Hence, $\frac{△}{15}=5$.