Q. If the area of the quadrilateral 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 contract of these tangents to the circle is 8 sq unit, 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) $c^{2} + 34c - 64 = 0$
Explanation
(b)
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 quadrilateral
$OACB = 2 \times area\ of\ the\ \mathrm{\Delta}OAC$
$= 2 \times \left( \frac{1}{2} \right) \times OA \times AC$
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)
$\Rightarrow \ \ c(34 - c) = 34$
$\Rightarrow c^{2} - 34c + 64 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.