# Circle A has a center at (3 ,4 ) and an area of 16 pi. Circle B has a center at (8 ,1 ) and an area of 40 pi. Do the circles overlap?

Mar 21, 2016

Yes. In fact the centre of circle A is contained in circle B.

#### Explanation:

The area of a circle is $\pi {r}^{2}$ where $r$ is the radius.

So the radius of circle A is $\sqrt{16} = 4$ and the radius of circle B is $\sqrt{40} = 2 \sqrt{10} \approx 6.325$

The distance between $\left(3 , 4\right)$ and $\left(8 , 1\right)$ is given by the distance formula as:

$\sqrt{{\left(8 - 3\right)}^{2} + {\left(1 - 4\right)}^{2}} = \sqrt{{5}^{2} + {3}^{2}} = \sqrt{25 + 9} = \sqrt{34} \approx 5.831$

So the point $\left(3 , 4\right)$ is actually contained in the circle B, being closer to $\left(8 , 1\right)$ than the radius of circle B.

graph{((x-3)^2+(y-4)^2-16)((x-3)^2+(y-4)^2-0.02)((x-8)^2+(y-1)^2-40)((x-8)^2+(y-1)^2-0.02) = 0 [-14.58, 25.42, -8.4, 11.6]}