# Circle A has a center at (5 ,4 ) and an area of 4 pi. Circle B has a center at (2 ,8 ) and an area of 9 pi. Do the circles overlap? If not, what is the shortest distance between them?

Nov 29, 2016

The circles touch, but they do not overlap. The shortest distance between them is $0$.

#### Explanation:

The distance between the points $\left(5 , 4\right)$ and $\left(2 , 8\right)$ is
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
$d = \sqrt{{\left(2 - 5\right)}^{2} + {\left(8 - 4\right)}^{2}}$
$d = \sqrt{{\left(- 3\right)}^{2} + {4}^{2}}$
$d = \sqrt{9 + 16}$
$d = \sqrt{25} = 5$

If the circles overlap, the sum of their radii will be greater than $d = 5.$ If they don't, the shortest distance between them will be what we get when we subtract both radii from $5$.

Let ${r}_{1}$ and ${r}_{2}$ be the radius of circles 1 and 2, respectively. Then:

${A}_{1} = \pi {r}_{1}^{2}$
$\implies {r}_{1} = \sqrt{{A}_{1} / \pi} = \sqrt{\frac{4 \cancel{\pi}}{\cancel{\pi}}} = \sqrt{4} = 2$

Similarly,
${r}_{2} = \sqrt{{A}_{2} / \pi} = \sqrt{\frac{9 \cancel{\pi}}{\cancel{\pi}}} = \sqrt{9} = 3$

We get
$\text{ } {r}_{1} + {r}_{2}$
$= 2 + 3$
$= 5 \text{ } = d$

Since the sum of the radii matches the distance between the two circles' centers, the circles touch, but they do not overlap. The shortest distance between the two circles is $0$.