# How do you find the equation of the circle with center at (-3, 1) and through the point (2, 13)?

${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = 169$

#### Explanation:

Let the equation of unknown circle with center $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(- 3 , 1\right)$ & radius $r$ be as follows

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

${\left(x - \left(- 3\right)\right)}^{2} + {\left(y - 1\right)}^{2} = {r}^{2}$

${\left(x + 3\right)}^{2} + {\left(y - 1\right)}^{2} = {r}^{2}$

Since, the above circle passes through the point $\left(2 , 13\right)$ hence it will satisfy the equation of circle as follows

${\left(2 + 3\right)}^{2} + {\left(13 - 1\right)}^{2} = {r}^{2}$

${r}^{2} = 25 + 144 = 169$

setting ${r}^{2} = 169$, we get the equation of circle

${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = 169$