# How do you find the equation of the circle with center (1,2) and radius 3?

Nov 10, 2016

${x}^{2} + {y}^{2} - 2 x - 4 y - 4 = 0$

#### Explanation:

$c e n t r e \left(1 , 2\right) , r a \mathrm{di} u s 3$

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

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

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = {3}^{2}$

multiplying out

${x}^{2} - 2 x + 1 + {y}^{2} - 4 y + 4 = 9$

tidying up

${x}^{2} + {y}^{2} - 2 x - 4 y - 4 = 0$