Question:
Let $A$ be the center of the cricle $x^2 + y^2 – 2x-4y-20=0$. Suppose that the tangents at the points $B(1,7)$ and $D(4,-2)$ on the cricle meet at point $C$. Find the area of the quadrilateral $ABCD$.
What I have done:
Well I have found the center of the circle and its radius. Upon drawing the diagram, it is obvious that the quadrilateral formed can be split into two right angled triangles. The only thing I need is the distance between the point of contact and the point of intersection of the tangents. How would I obtain this?
Best Answer
Hint. You know the lengths $AB$ and $AD$, and you know that $\angle ABC$ and $\angle ADC$ are right angles. Now find $\angle CAB=\frac{1}{2}\angle DAB$.
Alternative. The radius $AB$ is vertical, so the tangent $BC$ is horizontal, so $C$ has coordinates $(1+x,7)$ for some $x$. The area will be $5x$. Can you see how to find $x$ by using the fact that distances $BC$ and $DC$ are equal?