# What is the equation of the circle with a center at (7 ,1 ) and a radius of 2 ?

##### 1 Answer
Apr 24, 2016

$y = \pm$ \sqrt(4-(x²-14x+49))+1.

#### Explanation:

For a circle with center $\left(h , k\right)$ and radius $r$:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$.

So

${\left(x - 7\right)}^{2} + {\left(y - 1\right)}^{2} = 4$

${x}^{2} - 14 x + 49 + {y}^{2} - 2 y + 1 = 4$

${\left(y - 1\right)}^{2} = 4 - \left({x}^{2} - 14 x + 49\right)$

$\left(y - 1\right) = \setminus \sqrt{4 - \left({x}^{2} - 14 x + 49\right)}$

graph{(x-7)^2+(y-1)^2 = 4 [-1.42, 11.064, -2.296, 3.944]}