1. Let the point be $P(x,y)$.

2. Given condition:

   • Distance of $P$ from $(5,0)$ is thrice its distance from $(-5,0)$.

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

3. Squaring both sides: $(x-5)^2+y^2 = 9\left((x+5)^2+y^2\right)$

4. Expand both sides: $x^2-10x+25+y^2 = 9(x^2+10x+25+y^2)$

   $x^2-10x+25+y^2 = 9x^2+90x+225+9y^2$

5. Bring all terms to one side: $0 = 8x^2+100x+200+8y^2$

   Divide by $8$: $x^2+y^2+\frac{25}{2}x+25=0$

6. Compare with general circle form: $x^2+y^2+2gx+2fy+c=0$

   Here, $2g=\frac{25}{2}, \quad 2f=0, \quad c=25$

   So, $g=\frac{25}{4}, \quad f=0$

7. Radius of the circle: $r=\sqrt{g^2+f^2-c}$

   $r^2=\left(\frac{25}{4}\right)^2-25=\frac{625}{16}-\frac{400}{16}=\frac{225}{16}$

   Hence, $r=\frac{15}{4}$

8. Therefore, $4r^2 = 4\cdot \frac{225}{16} = \frac{225}{4}$

This is not an integer, so let us recheck the algebra carefully.

From step 5: $x^2-10x+25+y^2 = 9x^2+90x+225+9y^2$

Subtract LHS from RHS: $0=8x^2+100x+200+8y^2$

Divide by $4$ instead: $2x^2+25x+50+2y^2=0$

Now divide by $2$: $x^2+y^2+\frac{25}{2}x+25=0$

This is correct.

Complete the square: $x^2+\frac{25}{2}x+y^2+25=0$

$\left(x+\frac{25}{4}\right)^2-\frac{625}{16}+y^2+25=0$

$\left(x+\frac{25}{4}\right)^2+y^2=\frac{625}{16}-25=\frac{225}{16}$

So indeed, $r^2=\frac{225}{16}$

Thus, $4r^2=\frac{225}{4}$

But since the locus of a point dividing distances from two fixed points in ratio $3:1$ should be an Apollonius circle, let us verify whether the ratio was interpreted correctly.

Given: distance from $(5,0)$ is thrice distance from $(-5,0)$: $PF_1 = 3PF_2$ This leads exactly to the same equation above.

Hence the mathematically derived value is $4r^2=\frac{225}{4}$ not $56$.

So the stored answer appears inconsistent with the given question.