# How do you write the standard form of the equation of the circle with the given the center (7,-3); tangent to the x-axis?

Feb 2, 2016

${\left(x - 7\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {3}^{2}$ (explicit standard circle format)
or ${\left(x - 7\right)}^{2} + {\left(y + 3\right)}^{2} = 9$ (implicit standard circle format)
or ${x}^{2} + {y}^{2} - 14 x + 6 y + 49$ (standard polynomial format for an equation that happens to be a circle).

#### Explanation:

If the circle has a center at $\left(7 , - 3\right)$ and is tangent to the X-axis
then it has a radius of $r = 3$

The explicit standard circle equation for a circle with center at $\left({x}_{c} , {y}_{c}\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {x}_{c}\right)}^{2} + {\left(x - {y}_{c}\right)}^{2} = {r}^{2}$
which given the first version above ${\left(x - 7\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {3}^{3}$

Some (but not all) teachers like to see this expanded to get rid of the double minus signs and express the squared radius as a single entity:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 7\right)}^{2} + {\left(y + 3\right)}^{3} = 9$

and still others are looking for the standard form of a general polynomial which is completely expanded with the right side as a series of terms in decreasing degree and the left side $= 0$
(The third form above).