# How do you find the equation of the circle given center (2, 7) and its radius is 36?

Jun 17, 2016

${x}^{2} + {y}^{2} - 4 x - 14 y - 1243 = 0.$
The eqn. of a circle with centre at pt. $\left(h , k\right)$ & radius $= r$ is
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2.}$
Hence, the reqd. eqn. is ${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} = {36}^{2.}$
Upon simplification, it becomes ${x}^{2} + {y}^{2} - 4 x - 14 y - 1243 = 0.$