# Circle A has a center at (5 ,2 ) and an area of 15 pi. Circle B has a center at (2 ,1 ) and an area of 90 pi. Do the circles overlap?

##### 1 Answer
Oct 8, 2017

See a solution process below:

#### Explanation:

To find if the circles overlap we first must find the radius of each circle.

The formula for the area of a circle is:

$A = \pi {r}^{2}$

We can substitute for $A$ and solve for $r$:

Circle A:

$15 \pi = \pi {r}_{a}^{2}$

$\frac{15 \pi}{\textcolor{red}{\pi}} = \frac{\pi {r}_{a}^{2}}{\textcolor{red}{\pi}}$

$\frac{15 \textcolor{red}{\cancel{\textcolor{b l a c k}{\pi}}}}{\cancel{\textcolor{red}{\pi}}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\pi}}} {r}_{a}^{2}}{\cancel{\textcolor{red}{\pi}}}$

$15 = {r}_{a}^{2}$

$\sqrt{15} = \sqrt{{r}_{a}^{2}}$

$\sqrt{15} = {r}_{a}$

${r}_{a} = \sqrt{15}$

Circle B:

$90 \pi = \pi {r}_{b}^{2}$

$\frac{90 \pi}{\textcolor{red}{\pi}} = \frac{\pi {r}_{b}^{2}}{\textcolor{red}{\pi}}$

$\frac{90 \textcolor{red}{\cancel{\textcolor{b l a c k}{\pi}}}}{\cancel{\textcolor{red}{\pi}}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\pi}}} {r}_{b}^{2}}{\cancel{\textcolor{red}{\pi}}}$

$90 = {r}_{b}^{2}$

$\sqrt{90} = \sqrt{{r}_{b}^{2}}$

$\sqrt{90} = {r}_{b}$

${r}_{b} = \sqrt{90}$

Now, we can graph the two circles using the equation:

${\left(x - \textcolor{red}{a}\right)}^{2} + {\left(y - \textcolor{red}{b}\right)}^{2} = {\textcolor{b l u e}{r}}^{2}$

Where $\left(\textcolor{red}{a} , \textcolor{red}{b}\right)$ is the center of the circle and $\textcolor{b l u e}{r}$ is the radius of the circle.

Circle A:

${\left(x - \textcolor{red}{5}\right)}^{2} + {\left(y - \textcolor{red}{2}\right)}^{2} = {\textcolor{b l u e}{\sqrt{15}}}^{2}$

${\left(x - \textcolor{red}{5}\right)}^{2} + {\left(y - \textcolor{red}{2}\right)}^{2} = 15$

graph{((x-5)^2+(y-2)^2-15)=0 [-50. 50. -25. 25]}

Circle B:

${\left(x - \textcolor{red}{2}\right)}^{2} + {\left(y - \textcolor{red}{1}\right)}^{2} = {\textcolor{b l u e}{\sqrt{90}}}^{2}$

${\left(x - \textcolor{red}{2}\right)}^{2} + {\left(y - \textcolor{red}{1}\right)}^{2} = 90$

graph{((x-2)^2+(y-1)^2-90)((x-5)^2+(y-2)^2-15)=0 [-50. 50. -25. 25]}

The edges of the Circles DO NOT overlap.

However, Circle A is contained within Circle B