# Circle A has a radius of 2  and a center at (8 ,3 ). Circle B has a radius of 3  and a center at (4 ,5 ). If circle B is translated by <-3 ,4 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

##### 1 Answer
May 12, 2018

There's no overlap here. The closest points will be $\sqrt{85} - 5$ units apart.

#### Explanation:

Circle B's translated center is

$\left(4 , 5\right) + < - 3 , 4 > = \left(1 , 9\right)$

The distance $d$ between centers $\left(8 , 3\right)$ and $\left(1 , 9\right)$ is

$d = \sqrt{{\left(8 - 1\right)}^{2} + {\left(9 - 3\right)}^{2}} = \sqrt{85} \approx 9.2$

If the circles are to overlap, the distance between their centers must be less than or equal to the sum of the radii, $2 + 3 = 5.$ In other words, there needs to be a possible triangle whose sides ${r}_{A} , {r}_{B} \mathmr{and} d$ satisfy the triangle inequality. (Or almost satisfy, for here a degenerate triangle indicates the circles are tangent, which is acceptable overlap.) That would be a triangle with a vertex at each center and one at an intersection point.

So there's no overlap here. The closest points will be

$d - {r}_{A} - {r}_{B} = \sqrt{85} - 5$ units apart.

This was pretty easy, but sometimes when we're doing these sorts of problems we end up asking if the sum of two square roots is more or less than a third square root. Of course those can be done with a calculator, but there is a way to compare the squared lengths directly. It uses Archimedes' Theorem for the area $A$ of a triangle with sides $a , b , c ,$ which is a modern form of Heron:

$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2}$

A real triangle, one whose sides satisfy the triangle inequality, is one with a real area. So we require

$4 {a}^{2} {b}^{2} > {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2}$

or, when the degenerate triangle formed from three collinear points is acceptable,

$4 {a}^{2} {b}^{2} \ge {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2}$

This is all three constraints of the triangle inequality in a single inequality. We don't have to check $4 {a}^{2} {c}^{2}$ etc. separately.

In our example, ${a}^{2} = {r}_{A}^{2} = {2}^{2} = 4 ,$ ${b}^{2} = {r}_{B}^{2} = {3}^{2} = 9 ,$ and ${c}^{2} = {\left(8 - 1\right)}^{2} + {\left(9 - 3\right)}^{2} = 85.$ We check

 4(4)(9) stackrel{?}{ge} (85-4-9)^2

Nope, not a real or degenerate triangle, so no overlap.