Question

A point 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. The equation of locus of the point is
[a] $13{{x}^{2}}+13{{y}^{2}}-83x+64y+182=0$
[b] ${{x}^{2}}+{{y}^{2}}-11x+16y+26=0$
[c] ${{x}^{2}}+{{y}^{2}}-11x+16y=0$
[d] None of the above.

Answer 1

Hint: Assume that the coordinates of point be P(h,k). Using the distance formula find the distance between P and Q(3,-2). Also, using the formula for distance of a point from a line, find the distance of P from L: 5x-12y=13. Equate the square of PQ to the distance of P from L. Replace h by x and k by y to get the locus.

Complete step-by-step answer:
Let the coordinates of the point be P(h,k).
We know that the distance between the points $P\left( {{x}_{1}},{{y}_{1}} \right)$ and $Q\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $PQ=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$
Hence we have distance between P(h,k) and Q(3,-2) is given by \[PQ=\sqrt{{{\left( h-3 \right)}^{2}}+{{\left( k+2 \right)}^{2}}}\]
Also, we know that the distance of the point $P\left( {{x}_{1}},{{y}_{1}} \right)$ from the line $Ax+By+C=0$ is given by
$\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+C \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$
Hence the distance of P(h,k) from 5x-12y-13 = 0 is
$\dfrac{\left| 5h-12k-13 \right|}{\sqrt{{{5}^{2}}+{{12}^{2}}}}=\dfrac{\left| 5h-12k-13 \right|}{13}$
Hence we have
$\begin{align}
  & \dfrac{\left| 5h-12k-13 \right|}{13}={{\left( \sqrt{{{\left( h-3 \right)}^{2}}+{{\left( k+2 \right)}^{2}}} \right)}^{2}} \\
 & \Rightarrow \dfrac{\left| 5h-12k-13 \right|}{13}={{\left( h-3 \right)}^{2}}+{{\left( k+2 \right)}^{2}} \\
 & \Rightarrow 13\left( {{h}^{2}}-6h+9+{{k}^{2}}+4k+4 \right)=\left| 5h-12k-13 \right| \\
\end{align}$
Opening the modulo with +sign, we get
$13{{h}^{2}}+13{{k}^{2}}-83h+64k+182=0$
Opening the modulo with -ve sign, we get
$13{{h}^{2}}+13{{k}^{2}}-73h+40k+156=0$
Replacing h by x and k by y, we get
The loci of the points are $13{{x}^{2}}+13{{y}^{2}}-83x+64y+182=0$ and $13{{x}^{2}}+13{{y}^{2}}-73x+40y+156=0$
Hence option [a] is correct.

Note: [1] The above result shows that the locus of the point whose distance from a fixed line is square the distance from a fixed point is a circle.
[2] The circle $13{{x}^{2}}+13{{y}^{2}}-73x+40y+156=0$ is an imaginary circle because the quantity ${{g}^{2}}+{{f}^{2}}-c$ is negative for the given circle where g, f and c have their standard meaning.
[3] The whole circle should be present on one side of the line as we have chosen only one sign and neglected the other. This is evident from the graph below