Circle A has a center at (5 ,1 ) and an area of 23 pi. Circle B has a center at (2 ,8 ) and an area of 15 pi. Do the circles overlap?

Feb 6, 2017

Yes...

Explanation:

The area of a circle is given by the formula:

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

where $r$ is the radius.

So given the area $a$, the radius is given by the formula:

$r = \sqrt{\frac{a}{\pi}}$

Hence:

• Circle A has radius $\sqrt{23}$

• Circle B has radius $\sqrt{15}$

The distance $d$ between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by the formula:

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

So the distance between the centre $\left({x}_{1} , {y}_{1}\right) = \left(5 , 1\right)$ of circle A and the centre $\left({x}_{2} , {y}_{2}\right) = \left(2 , 8\right)$ of circle B is:

$d = \sqrt{{\left(2 - 5\right)}^{2} + {\left(8 - 1\right)}^{2}} = \sqrt{{3}^{2} + {7}^{2}} = \sqrt{9 + 49} = \sqrt{58}$

Then we find:

$\sqrt{23} + \sqrt{15} \approx 4.8 + 3.9 = 8.7 > 7.6 \approx \sqrt{58}$

Since the sum of the radii is greater than the distance between the centres of the circles, they do overlap...

graph{((x-5)^2+(y-1)^2-23)((x-5)^2+(y-1)^2-0.09)((x-2)^2+(y-8)^2-15)((x-2)^2+(y-8)^2-0.1) = 0 [-12, 22, -5, 12]}