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 the locus of the point is

• A. $13x^2+13y^2–83x+64y+182=0$
• B. $x^2+y^2–11x+16y+26=0$
• C. $x^2+y^2–11x+16y=0$
• D. None of these

Answer 1

The correct option is A

$13x^2+13y^2–83x+64y+182=0$

Finding the equation of the locus of the point:

Let be the moving point & given point be $A(3,-2)$

$\begin{array}{rcl}AP& =& \sqrt{{(h-3)}^2+{(k+2)}^2}\\ A{P}^2& =& {(h-3)}^2+{(k+2)}^2\end{array}$

Distance from the point $P(h,k)$from the line $5x-12y=13$ is

$D=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$

$\begin{array}{rcl}D& =& \frac{\left|5h-12k-13\right|}{\sqrt{5^2+{12}^2}}\\ & =& \frac{\left|5h-12k-13\right|}{13}\end{array}$

Given, $A{P}^2=D$

$$\begin{gathered} \Rightarrow {(h-3)}^2+{(k+2)}^2=\frac{\left|5h-12k-13\right|}{13} \\ \Rightarrow 13(h^2–6h+9+k^2+4k+4)=(5h–12k–13) \\ \Rightarrow 0=13h^2–78h+117+13k^2+52k+52-5h+12k+13 \\ \Rightarrow 13h^2+13k^2–83h+64k+182=0 \end{gathered}$$

So the locus of the point is

$13x^2+13y^2-83x+64y+182=0$

Hence, the correct option is$\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{.}$