# How do you write the equation for a circle with center at (4,1) and a radius of 3?

Jul 22, 2016

${x}^{2} + {y}^{2} - 8 x - 2 y + 8 = 0$

#### Explanation:

Suppose that the centre of a circle is $\left({x}_{0} , {y}_{0}\right)$ and, its radius $r$.

Then, eqn. of the circle is ${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$

Therefore, the reqd. eqn. is, ${\left(x - 4\right)}^{2} + {\left(y - 1\right)}^{2} = {3}^{2}$, i.e.,

${x}^{2} + {y}^{2} - 8 x - 2 y + 8 = 0$