# How do you write the equation for a circle with center at (2,3) that is tangent to the x-axis?

${\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 9$

#### Explanation:

Given that the center of circle is $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(2 , 3\right)$.

The above circle is tangent to the x-axis hence its radius

$r = \setminus \textrm{y - c \infty r \mathrm{di} n a t e} = 3$

hence the equation of the circle is given by following formula

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

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

${\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 9$

Jul 26, 2018

${\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 9$

#### Explanation:

$\text{the equation of a circle in standard form is}$

•color(white)(x)(x-a)^2+(y-b)^2=r^2

$\text{where "(a,b)" are the coordinates of the centre and r}$
$\text{is the radius}$

$\text{here } \left(a , b\right) = \left(2 , 3\right)$

$\text{r is the vertical distance from the x-axis to the centre}$

$r = \text{y-coordinate of centre } = 3$

${\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 9 \leftarrow \textcolor{red}{\text{equation of circle}}$