# Circle A has a center at (3 ,5 ) and an area of 78 pi. Circle B has a center at (1 ,2 ) and an area of 54 pi. Do the circles overlap?

May 9, 2018

Yes

#### Explanation:

First, we need the distance between the two centres, which is $D = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}}$

$D = \sqrt{{\left(5 - 2\right)}^{2} + {\left(3 - 1\right)}^{2}} = \sqrt{{3}^{2} + {2}^{2}} = \sqrt{9 + 4} = \sqrt{13} = 3.61$

Now we need the sum of radii, since:
D>(r_1+r_2);"Circles don't overlap"
D=(r_1+r_2);"Circles just touch"
D<(r_1+r_2);"Circles do overlap"

pir_1""^2=78pi
r_1""^2=78
${r}_{1} = \sqrt{78}$

pir_2""^2=54pi
r_2""^2=54
${r}_{2} = \sqrt{54}$

$\sqrt{78} + \sqrt{54} = 16.2$

$16.2 > 3.61$, so circles do overlap.

Proof:
graph{((x-3)^2+(y-5)^2-54)((x-1)^2+(y-2)^2-78)=0 [-20.33, 19.67, -7.36, 12.64]}

May 10, 2018

These overlap if $\sqrt{78} + \sqrt{54} \ge \sqrt{{\left(3 - 1\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{13} .$

We can skip the calculator and check $4 \left(13\right) \left(54\right) \ge {\left(78 - 13 - 54\right)}^{2}$ or $4 \left(13\right) \left(54\right) > {11}^{2}$ which it surely is, so yes, overlap.

#### Explanation:

Circle area is of course $\pi {r}^{2}$ so we divide out the gratuitous $\pi$s.

${r}_{1}^{2} = 78$

${r}_{2}^{2} = 54$

and squared distance between the centers

${d}^{2} = {\left(3 - 1\right)}^{2} + {\left(5 - 2\right)}^{2} = 13$

Basically we want to know if ${r}_{1} + {r}_{2} \ge d$, i.e. if we can make a triangle out of two radii and the segment between the centers.

The squared lengths are all nice integers and it's pretty insane that we all instinctively reach for the calculator or computer and start taking square roots.

We don't have to, but it requires a little detour. Let's use Heron's formula, call the area $Q$.

$Q = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ where $s = \frac{a + b + c}{2}$

${Q}^{2} = \left(\frac{a + b + c}{2}\right) \left(\left(\frac{a + b + c}{2}\right) - a\right) \left(\left(\frac{a + b + c}{2}\right) - b\right) \left(\left(\frac{a + b + c}{2}\right) - c\right)$

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

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

That's already better than Heron. But we continue. I'll skip some tedium.

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

That's nicely symmetric, as we would expect for an area formula. Let's make it less symmetric looking. Recall

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

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

That's a formula for the squared area of a triangle given the squared lengths of the sides. When the latter are rational, so is the former.

Let's try it out. We're free to assign the sides however we like; for hand calculation its best to make $c$ the largest side,

${c}^{2} = 78$

${a}^{2} = 54$

${b}^{2} = 13$

$16 {Q}^{2} = 4 \left(54\right) \left(13\right) - {\left(78 - 54 - 13\right)}^{2} = 4 \left(54\right) 13 - {11}^{2}$

Even before calculating it any more, we can see we have a positive $16 {Q}^{2}$ so a real triangle with a positive area, so overlapping circles.

$16 {Q}^{2} = 2687$

If we had gotten an negative value, an imaginary area, that's not a real triangle, so non-overlapping circles.