Q. If the lines $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$ are two diameters of a circle of area $49\pi$ sq unit, the equation of the circle is
A) $x^{2} + y^{2} + 2x - 2y - 62 = 0$
B) $x^{2} + y^{2} - 2x + 2y - 62 = 0$
C) $x^{2} + y^{2} - 2x + 2y - 47 = 0$
D) $x^{2} + y^{2} + 2x - 2y - 47 = 0$
Explanation
(c)
$\therefore$ The intersection of two diameter is the centre of circle, is $(1,\ - 1)$
Let $r$ be the radius of circle, then
$\Rightarrow \$ Area of circle $\pi r^{2} = 49\pi\ \ \Rightarrow \ \ r = 7$ unit
$\therefore$ Equation of required circle is
$(x - 1)^{2} + (y + 1)^{2} = 49$
$\Rightarrow \ \ \ \ x^{2} + y^{2} - 2x + 2y - 47 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.