# How do you write an equation of a circle given center at the point (1, 0) and has the point (-2, 3)?

Mar 20, 2016

${x}^{2} + {y}^{2} - 2 x - 17 = 0$

#### Explanation:

The equation of the circle with center at (a, b) and radius r is
${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$.

Here a = 1 and b = 0.
The equation becomes
${\left(x - 1\right)}^{2} + {y}^{2} = {r}^{2}$.

The circle passes through ($- 2$, 3).
So, ${r}^{2}$ = 18.
Answer .${\left(x - 1\right)}^{2} + {y}^{2} = 18$ or .${x}^{2} + {y}^{2} - 2 x - 17 = 0$