Question

Let a point $P$ be such that its distance from the point $(5,0)$ is thrice the distance of $P$ from the point $(-5,0)$. If the locus of the point $P$ is a circle of radius $r$, then $4r^2$ is equal to ?

Answer 1

Finding the locus of point $P$

Considering the points according to the given data:

$P(h,k),A(5,0)$ and $B(-5,0)$

Given, $PA=3PB$

Distance between the points $P(h,k)\&A(5,0)$ is:

$PA=\sqrt{{(h-5)}^2+k^2}$

Similarly, the distance between the points $P(h,k)\&B(-5,0)$

$PB=\sqrt{{(h+5)}^2+k^2}$

Equating the two equations:

$$\begin{gathered} PA=3PB\left[Given\right] \\ \Rightarrow P{A}^2=9P{B}^2[squaringbothsides] \\ \Rightarrow {(h-5)}^2+k^2=9[{(h+5)}^2+k^2] \\ \Rightarrow 8h^2+8k^2+100h+200=0 \\ \Rightarrow x^2+y^2+\frac{25}{2}x+25=0[Dividingbothsidesby8] \end{gathered}$$

Therefore, the locus of point $P$ is $x^2+y^2+\frac{25}{2}x+25=0.....\left(i\right)$

The above equation also represents the equation of the circle.

Finding the radius of the circle:

We know that a circle of the form $x^2+y^2+2gx+2fy+c=0$has a center $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$.

Therefore, from equation $\left(i\right)$ we have,

$g=\frac{25}{4},f=0\&c=25$

Thus, Centre $=\left(-g,-f\right)=\left(-\frac{25}{4},0\right)$

Calculating the radius:

$$\begin{gathered} \Rightarrow r=\sqrt{{\left(\frac{25}{4}\right)}^2-25} \\ \Rightarrow r=\sqrt{\frac{625}{16}-25} \\ \Rightarrow r=\sqrt{\frac{225}{16}} \\ \Rightarrow r^2=\frac{225}{16}[squaringbothsides] \\ \Rightarrow 4r^2=\frac{225}{4}[multiplyingwith4bothsides] \\ \Rightarrow 4r^2=56.25 \end{gathered}$$

Hence, the correct answer is $56.25$