# What is the standard form of the equation of a circle with center (-3,3) and tangent to the line y=1?

May 7, 2016

Equation of circle is ${x}^{2} + {y}^{2} + 6 x - 6 y + 14 = 0$ and $y = 1$ is tangent at $\left(- 3 , 1\right)$

#### Explanation:

The equation of a circle with center $\left(- 3 , 3\right)$ with radius $r$ is

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

or ${x}^{2} + {y}^{2} + 6 x - 6 y + 9 + 9 - {r}^{2} = 0$

As $y = 1$ is a tangent to this circle, putting $y = 1$ in the equation of a circle should give only one solution for $x$. Doing so we get

${x}^{2} + 1 + 6 x - 6 + 9 + 9 - {r}^{2} = 0$ or

${x}^{2} + 6 x + 13 - {r}^{2} = 0$

and as we should have only one solution, discriminant of this quadratic equation should be $0$.

Hence, ${6}^{2} - 4 \times 1 \times \left(13 - {r}^{2}\right) = 0$ or

$36 - 52 + 4 {r}^{2} = 0$ or $4 {r}^{2} = 16$ and as $r$ has to be positive

$r = 2$ and hence equation of circle is

${x}^{2} + {y}^{2} + 6 x - 6 y + 9 + 9 - 4 = 0$ or ${x}^{2} + {y}^{2} + 6 x - 6 y + 14 = 0$

and $y = 1$ is tangent at $\left(- 3 , 1\right)$