# How would you find the equation of the circle, center at (5,-3) and radius of 4?

Mar 4, 2018

${x}^{2} + {y}^{2} - 10 x + 6 y + 18 = 0$

#### Explanation:

the eqn of a circle with centre $\left(a , b\right)$ radius $r$

is given by

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

in this case we have

centre $\left(5 , - 3\right)$, radius$\text{ } r = 4$

eqn

${\left(x - 5\right)}^{2} + {\left(y - - 3\right)}^{2} = {4}^{2}$

$\implies {\left(x - 5\right)}^{2} + {\left(y + 3\right)}^{2} = {4}^{2}$

expanding

${x}^{2} - 10 x + 25 + {y}^{2} + 6 y + 9 = 16$

${x}^{2} + {y}^{2} - 10 x + 6 y + 18 = 0$