Question

The locus of the point which moves so that the square of its distance from the point $(3,-2)$ is numerically equal to its distance from the line $5x-12y=13$ can be

• A. $x^2+y^2+14+\frac{83x-64y}{13}=0$
• B. $x^2+y^2+14=\frac{83x-64y}{13}$
• C. $x^2+y^2+12+\frac{73x-40y}{13}=0$
• D. $x^2+y^2+12=\frac{73x-40y}{13}$

Answer 1

Answer: (B), (D)

$x^2+y^2+12=\frac{73x-40y}{13}$

Let the variable point be $(h,k)$
According to the given condition, we get
$(h-3{)}^2+(k+2{)}^2=\frac{|5h-12k-13|}{13}$

Now
Case $1$: Point $(h,k)$ and origin lies on the same side of the line $5x-12y-13=0$
$(h-3{)}^2+(k+2{)}^2=-\frac{(5h-12k-13)}{13}\Rightarrow 13(h^2+k^2)+169-78h+52k=-5h+12k+13\Rightarrow 13(h^2+k^2+12)-73h+40k=0$
Therefore, the locus is
$x^2+y^2+12=\frac{73x-40y}{13}$

Case $2$: Point $(h,k)$ and origin lies on opposite sides of the line $5x-12y-13=0$
$(h-3{)}^2+(k+2{)}^2=\frac{(5h-12k-13)}{13}\Rightarrow 13(h^2+k^2)+169-78h+52k=5h-12k-13\Rightarrow 13(h^2+k^2+14)-83h+64k=0$
Therefore, the locus is
$x^2+y^2+14=\frac{83x-64y}{13}$