# How do you find general form of circle with Center at the point (1, 0) and having the point (-2, 3)?

Dec 4, 2015

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

#### Explanation:

The general form for a circle with center $\left(a , b\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

If the center of the circle is at $\left({x}_{c} , {y}_{c}\right) = \left(1 , 0\right)$
and
$\left({x}_{p} , {y}_{p}\right) = \left(- 2 , 3\right)$ is a point on the circle, then the radius squared is
$\textcolor{w h i t e}{\text{XXX")r^2 = (x_p-x_c)^2+(y_p-y_c)^2color(white)("XXX}}$(Pythagorean Theorem)

$\textcolor{w h i t e}{\text{XXXX}} = {\left(- 2 - 1\right)}^{2} + {\left(3 - 0\right)}^{2} = {3}^{2} + {3}^{2} = 9 + 9 = 18$

Combining this information concerning the radius (squared)
and the general form for a circle, we get
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 1\right)}^{2} + {\left(y - 0\right)}^{2} = 18$