Q. A point moves in such a way that the square of its distance from the point $(3,\ - 2)$ is equal to numerically its distance from the line $5x - 12y = 13$. The equation of the locus of the point is
A) $x^{2} + y^{2} - 11x - 16y + 26 = 0$
B) $x^{2} + y^{2} - 11x + 16y = 0$
C) $13\left( x^{2} + y^{2} \right) - 83x + 64y + 182 = 0$
D) $x^{2} + y^{2} - 83x + 64y + 182 = 0$
Explanation
(c)
Let $(h,\ k\ )$ be the point such that
$(h - {3)}^{2} + (k + 2)^{2} = \frac{5h - 12k - 13}{\sqrt{25 + 144}}$
$\Rightarrow 13\left( h^{2} + 9 - 6h + k^{2} + 4k + 4 \right) = 5h - 12k - 13$
$\Rightarrow 13\left( h^{2} + k^{2} \right) - 83h + 64k + 182 = 0$
Thus, the locus of $(h,\ k)\$is
$13\left( x^{2} + y^{2} \right) - 83x + 64y + 182 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.