Q. If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circle of radius 7, then the equation of the circle is
A) $x^{2} + y^{2} + 2x - 4y - 47 = 0$
B) $x^{2} + y^{2} = 49$
C) $x^{2} + y^{2} - 2x + 2y - 47 = 0$
D) $x^{2} + y^{2} = 17$
Explanation
(c)
The intersection of diameter lines is the centre of the circle, $ie,\ \ C(1,\ - 1)$
$\therefore$ Required equation of circle is
$(x - 1)^{2} + (y + 1)^{2} = 7^{2}$
$\Rightarrow \ x^{2} + y^{2} - 2x + 2y - 47 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.