# What is the standard form of the equation of a circle with a center(1,-2) and passes through (6,-6)?

Jul 9, 2018

The circle equation in standard form is

${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$

Where (x_0,y_0); r are the center coordinates and radius

We know that $\left({x}_{0} , {y}_{0}\right) = \left(1 , - 2\right)$, then

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

But we know that passes trough $\left(6 , - 6\right)$, then

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

${5}^{2} + {\left(- 4\right)}^{2} = 41 = {r}^{2}$, So $r = \sqrt{41}$

Finally we have the standard form of this circle

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

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

#### Explanation:

Let the equation of unknown circle with center $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(1 , - 2\right)$ & radius $r$ be as follows

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

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

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

Since, the above circle passes through the point $\left(6 , - 6\right)$ hence it will satisfy the equation of circle as follows

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

${r}^{2} = 25 + 16 = 41$

setting ${r}^{2} = 41$, we get the equation of circle

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