Question

If the area of the quadrilateral formed by the tangent from the origin to the circle $x^2+y^2+6x-10y+c=0$ and the pair of radii at the points of contact of these tangents to tbe circle is $8$ square units. then $c$ is a root of the equation

• A. $c^2-32c+64=0$.
• B. $c^2-34c+64=0$.
• C. $c^2+2c-64=0$.
• D. $c^2+34c-64=0$.

Answer 1

Answer: (B)

$c^2-34c+64=0$.

Let $OA,OB$ be the tangents from the origin to the given circle with centre $C(-3,5)$
and radius .$\sqrt{9+25-c}=\sqrt{34-c}$
Then area of the quadribiteral $OACB=2\times$ area of $△OAC=2\times \left(\frac{1}{2}\right)\times OA\times AC$
Now $OA=$ length of the tangent from the origin to the given circle $=.\sqrt{C}$
and $AC=$ radius of the circle $=\sqrt{34-c}$ so that. $\sqrt{C}\sqrt{34-c}=8$ ...(given)
$\Rightarrow c(34-c)=64\Rightarrow c^2-34c+64=0$