# Circle A has a center at (3 ,2 ) and an area of 96 pi. Circle B has a center at (12 ,7 ) and an area of 48 pi. Do the circles overlap?

May 6, 2016

Both circles overlap.

#### Explanation:

Circle $A$, Area$= \pi {r}^{2} = 96 \pi$
Solving for radius $r$,
$\pi {r}^{2} = 96 \pi$
or ${r}^{2} = 96$
or $r = \sqrt{96}$
${r}_{A} = 4 \sqrt{6}$
${r}_{A} = 9.8$, rounded to one decimal place
Similarly Circle $B$
Area$= \pi {r}^{2} = 48 \pi$
Solving for radius $r$,
$\pi {r}^{2} = 48 \pi$
or ${r}^{2} = 48$
or $r = \sqrt{48}$
${r}_{B} = 4 \sqrt{3}$
${r}_{B} = 4 \sqrt{3}$
${r}_{B} = 6.4$, rounded to one decimal place
Now ${r}_{A} + {r}_{B} = 9.8 + 6.4 = 16.2$, rounded to one decimal place

Distance between the two centers, $\left(3 , 2\right) \mathmr{and} \left(12 , 7\right)$
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
or $d = \sqrt{{\left(12 - 3\right)}^{2} + {\left(7 - 2\right)}^{2}}$
or $d = \sqrt{{\left(9\right)}^{2} + {\left(5\right)}^{2}}$
or $d = \sqrt{106}$
or $d = 10.3$, rounded to one decimal place

Since the sum of both the radii is $16.2 > d$, the distance between the two centers.
Therefore, both the circles overlap.