# How do you find the equation of the circle passing through (7,5) and (3,7), and with center on x-3y+3=0?

${\left(x - 3\right)}^{2} + {\left(y - 2\right)}^{2} = 25$

#### Explanation:

Let $\left({x}_{1} , {y}_{1}\right)$ be the center & $r$ be the radius of unknown circle then as per given conditions, the distance of each of points $\left(7 , 5\right)$ & $\left(3 , 7\right)$ from center $\left({x}_{1} , {y}_{1}\right)$ will be equal to radius $r$ as follows

${\left({x}_{1} - 7\right)}^{2} + {\left({y}_{1} - 5\right)}^{2} = {r}^{2} \setminus \ldots \ldots . \left(1\right)$

${\left({x}_{1} - 3\right)}^{2} + {\left({y}_{1} - 7\right)}^{2} = {r}^{2} \setminus \ldots \ldots . \left(2\right)$

Subtracting (1) from (2), we get

$2 {x}_{1} - {y}_{1} - 4 = 0 \setminus \ldots \ldots \ldots \ldots . . \left(3\right)$

Since, the center $\left({x}_{1} , {y}_{1}\right)$ of circle lies on the straight line: $x - 3 y + 3 = 0$ hence it will satisfy the equation f straight line as follows

${x}_{1} - 3 {y}_{1} + 3 = 0 \setminus \ldots \ldots . . \left(4\right)$

solving (3) & (4), we get ${x}_{1} = 3 , {y}_{1} = 2$

substituting the values of ${x}_{1} = 3 , {y}_{1} = 2$ in (1), we get ${r}^{2} = 25$

hence, the equation of circle is

${\left(x - {x}_{1}\right)}^{2} + {\left(y - {y}_{1}\right)}^{2} = {r}^{2}$

${\left(x - 3\right)}^{2} + {\left(y - 2\right)}^{2} = 25$