# Circle A has a center at (6 ,5 ) and an area of 6 pi. Circle B has a center at (12 ,7 ) and an area of 48 pi. Do the circles overlap?

May 10, 2018

Since

${\left(12 - 6\right)}^{2} + {\left(7 - 5\right)}^{2} = 40 \quad$ and

$4 \left(6\right) \left(48\right) - {\left(40 - 6 - 48\right)}^{2} = 956 > 0$

we can make a real triangle with squared sides 48, 6 and 40, so these circles intersect.

#### Explanation:

Why the gratuitous $\pi$?

The area is $A = \pi {r}^{2}$ so ${r}^{2} = \frac{A}{\pi} .$ So the first circle has a radius ${r}_{1} = \setminus \sqrt{6}$ and the second ${r}_{2} = \sqrt{48} = 4 \sqrt{3}$.

The centers are $\sqrt{{\left(12 - 6\right)}^{2} + {\left(7 - 5\right)}^{2}} = \sqrt{40} = 2 \sqrt{10}$ apart.

So the circles overlap if $\sqrt{6} + 4 \sqrt{3} \ge 2 \sqrt{10}$.

That's so ugly that you'd be forgiven for reaching for the calculator. But it's really not necessary. Let's take a detour and look how this is done using Rational Trigonometry. There we're only concerned with the squared lengths, called quadrances.

Let's say we want to test if three quadrances $A , B , C$ are the quadrances between three collinear points, i.e. $\sqrt{A} = \sqrt{B} + \sqrt{C}$ or $\sqrt{B} = \sqrt{A} + \sqrt{C} ,$ or $\sqrt{C} = \sqrt{A} + \sqrt{B}$. We'll write it as

$\pm \sqrt{C} = \pm \sqrt{A} \pm \sqrt{B}$

Squaring,

$C = A + B \setminus \pm 2 \sqrt{A B}$

$C - A - B = \pm 2 \sqrt{A B}$

Squaring again,

${\left(C - A - B\right)}^{2} = 4 A B$

$0 = 4 A B - {\left(C - A - B\right)}^{2}$

It turns out

$m a t h c a l \left\{A\right\} = 4 A B - {\left(C - A - B\right)}^{2}$

is a discriminant for triangles. We just showed if $m a t h c a l \left\{A\right\} = 0$ that means we have a degenerate triangle, formed from three collinear points. If $m a t h c a l \left\{A\right\} > 0$ then we have a real triangle, each side less than the sum of the other two. If $m a t h c a l \left\{A\right\} < 0$ we don't have sides that satisfies the triangle inequality, and we sometimes call this an imaginary triangle.

Let's turn back to our question armed with our new triangle discriminant $m a t h c a l \left\{A\right\}$. If the circles intersect we we can make a triangle of the two centers and an intersection, so the sides will have lengths ${r}_{1}$, ${r}_{2}$, and the distance between the centers $\left(6 , 5\right)$ and $\left(12 , 7\right)$. We have

$A = {r}_{1}^{2} = 6$

$B = {r}_{2}^{2} = 48$

$C = {\left(12 - 6\right)}^{2} + {\left(7 - 5\right)}^{2} = 40$

$m a t h c a l \left\{A\right\} = 4 A B - {\left(C - A - B\right)}^{2} = 4 \left(6\right) \left(48\right) - {\left(40 - 6 - 48\right)}^{2} = 956$

$m a t h c a l \left\{A\right\} > 0$ so we have a real triangle, i.e overlapping circles.

Oh yeah, for any triangle $m a t h c a l \left\{A\right\} = 16 {\left(\textrm{a r e a}\right)}^{2} .$

Check: Alpha