# How do you find the general form of the equation of this circle given Center (2,3) and tangent to the x-axis?

May 1, 2018

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

#### Explanation:

If the centre is at (2,3) and the x axis is a tangent to the circle then the radius must be 3.

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ is the general formula for a circle with centre $\left(a , b\right)$ and radius $r$

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

May 1, 2018

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

#### Explanation:

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{with centre "=(-g,-f)" and}$

"and radius "=sqrt(g^2+f^2-c)color(white)(x);g^2+f^2-c>0

$\text{to begin obtain the equation in standard form}$

•color(white)(x)(x-a)^2+(y-b)^2=r^2

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

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

$\text{and "r=3to"(distance from x-axis to y-coord of centre)}$

$\Rightarrow {\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 9 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{distributing and rearranging}$

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

$\Rightarrow {x}^{2} + {y}^{2} - 4 x - 6 y + 4 = 0 \leftarrow \textcolor{red}{\text{in general form}}$