# Circle A has a center at (12 ,9 ) and an area of 16 pi. Circle B has a center at (3 ,1 ) and an area of 67 pi. Do the circles overlap?

Apr 27, 2018

Yes.

#### Explanation:

First we need to find the radii of the circles. We can do this using the formula for area:

"Area=pir^2

Circle A

$\pi {r}^{2} = 16 \pi$

${r}^{2} = 16 \implies r = \sqrt{16} = 4$

Circle B

$\pi {r}^{2} = 67 \pi$

${r}^{2} = 67 \implies r = \sqrt{67}$

We now find the distance between the centres. We can use the distance formula for this:

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

Coordinates $\left(12 , 9\right) , \left(3 , 1\right)$

$d = \sqrt{{\left(12 - 3\right)}^{2} + {\left(9 - 1\right)}^{2}} = \sqrt{145}$

Let $\setminus \setminus \setminus \setminus d = \text{distance between centres}$

and $\setminus \setminus \setminus s = \text{sum of radii}$

If:

$s = d \setminus \setminus \setminus \setminus \setminus$ circles touch at one point.

$s > d \setminus \setminus \setminus \setminus \setminus$ circles intersect at two points.

$s < d \setminus \setminus \setminus \setminus \setminus \setminus$circles do not touch.

$4 + \sqrt{67}$
$4 + \sqrt{67} > \sqrt{145}$