# How do you find general form of circle centered at (1,0) and containing point (-2,3)?

Dec 20, 2015

${\left(x - 1\right)}^{2} + {y}^{2} = 18$

#### Explanation:

The general form of a circle:

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

where $\left(h , k\right)$ is the center of the circle and $r$ is the radius.

Since the center of the circle is $\left(1 , 0\right)$, we can fill in $h$ and $k$.

${\left(x - 1\right)}^{2} + {\left(y - 0\right)}^{2} = {r}^{2}$

Now, all we need to know is the radius length, which we can find by applying to distance formula with the circle's center and another point on the circle: $\left(- 2 , 3\right)$.

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

$d = \sqrt{{\left(- 2 - 1\right)}^{2} + {\left(3 - 0\right)}^{2}}$

$d = \sqrt{18}$

Thus, the radius is $\sqrt{18}$, and the radius is represented in the general form of a circle by ${r}^{2}$, so ${r}^{2} = 18$.

Thus, the equation of the circle is:

${\left(x - 1\right)}^{2} + {y}^{2} = 18$

graph{(x-1)^2+y^2=18 [-10, 10, -5, 5]}