# How do you write the equation of the circle with center(1,-2) and passes through (6,-6)?

Apr 24, 2016

${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = 41$

#### Explanation:

The equation of a circle with centre $\left(h , k\right)$ and radius $r$ may be written:

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

We are given $\left(h , k\right) = \left(1 , - 2\right)$, so the only unknown is ${r}^{2}$.

Since the circle passes through $\left(6 , - 6\right)$, the values $x = 6$, $y = - 6$ will satisfy the equation and we find:

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

$= {\left(6 - 1\right)}^{2} + {\left(\left(- 6\right) - \left(- 2\right)\right)}^{2}$

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

$= 25 + 16$

$= 41$

So the equation of our circle may be written:

${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = 41$

graph{((x-1)^2+(y+2)^2-41)((x-1)^2+(y+2)^2-0.02)((x-6)^2+(y+6)^2-0.02) = 0 [-8.58, 11.42, -8.52, 1.48]}