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

Sep 8, 2016

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

#### Explanation:

The general form of equation of a circle is

${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$, whose center is $\left(- g , - f\right)$.

As the center is $\left(1 , 0\right)$, the equation would be of the form

x^2+y^2+2(-1)x+2×0×y+c=0 or

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

As it passes through $\left(- 2 , 3\right)$, putting these values in the equation we get $c$.

(-2)^2+3^2-2×(-2)+c=0 or

$4 + 9 + 4 + c = 0$ i.e. $c = - 17$.

Hence equation of circle is

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