Question

For a $△ABC$ the vertices are $A(0,3),B(0,12)$ and $C(x,0)$. If the circumcircle of $△ABC$ touches the $x$-axis, then the area (in sq. units) of the $△ABC$ is

• A. 36
• B. 27
• C. 30
• D. 24

Answer 1

Answer: (B)

$OA\times OB=O{C}^2$
$\Rightarrow 3\times 12=x^2$
$\Rightarrow x=6$
Centre $\left(C^{\prime }\right)=\left(6,\frac{15}{2}\right)$
$\angle A{C}^{\prime }B=2\angle ACB=2\theta$
Area of $△ACB=\frac{1}{2}\times 9\times 6=27$ sq. units