A moving point has its distance from (1,3) always one-third of its distance from (8,2). Find the equation of its Locus. A moving point has its distance from (1,3) always one-third of its distance from (8,2).
Find the equation of its Locus.
My equation displays a circle formed by the loci, I don't know if it's right.  Please help me find the equation, Thank you in advance to those that can answer.

$$\begin{align*}3^2\left[(x-1)^2+(y-3)^2\right] =& (x-8)^2 + (y-2)^2\\
9\left(x^2-2x+y^2-6y+10\right) =& x^2-16x+y^2-4x+68\\
8x^2-2x+8y^2-50y+22 =& 0\\
x^2-\frac14x+y^2-\frac{25}4y+\frac{11}4=&0\\
\left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\right)^2=&\frac{225}{32}\\
\left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\right)^2=&\left(\frac{15\sqrt2}{8}\right)^2
\end{align*}$$
