# How do you write the equation in standard form for a circle with center (-3,7) for a circle and tangent to the x axis?

Feb 2, 2016

${\left(x + 3\right)}^{2} + {\left(y - 7\right)}^{2} = 49$

#### Explanation:

The standard form is ${x}^{2} + {y}^{2} = {r}^{2}$. First, let's determine $r$. The center is $7$ above the x-axis and the circle is tangent to the x-axis, so the radius $r$ should be equal to $7$.

The center moved $3$ to the left, so substitute $x$ by $\left(x + 3\right)$.
The center moved $7$ up, so substitute $y$ by $\left(y - 7\right)$.
You can determine these numbers by filling in the center coördinates, the outcome must be zero. [$x + 3 = - 3 + 3 = 0$]
Maybe you've seen the standard equation ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ for a circle with center $\left(a , b\right)$.

So, the equation becomes ${\left(x + 3\right)}^{2} + {\left(y - 7\right)}^{2} = {7}^{2} = 49$