Solve this following


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 $\mathrm{r}$, then $4 \mathrm{r}^{2}$ is equal to



Jet noint is (h. k)

So, $\sqrt{(h-5)^{2}+k^{2}}=3 \sqrt{(h+5)^{2}+k^{2}}$

$8 x^{2}+8 y^{2}+100 x+200=0$

$x^{2}+y^{2}+\frac{25}{2} x+25=0$


$4 \mathrm{r}^{2}=\frac{25^{2}}{4}-100$

$4 r^{2}=156.25-100$

$4 r^{2}=56.25$

After round of $4 \mathrm{r}^{2}=56$


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