# Circle A has a center at (1 ,3 ) and an area of 16 pi. Circle B has a center at (2 ,7 ) and an area of 28 pi. Do the circles overlap?

Feb 19, 2016

The distance from the centre of circle A to the centre of circle B is less than the sum of the two radii. Thus we can conclude that the two circles are overlapping.

#### Explanation:

area of a circle = $\pi {r}^{2}$
area of circle A $= 16 \pi$
area of circle B $= 28 \pi$
radius of circle A = ${r}_{A} = \sqrt{\frac{16 \pi}{\pi}} = \sqrt{16} = 4$
radius of circle B $= {r}_{B} = \sqrt{\frac{28 \pi}{\pi}} = \sqrt{28} = 2 \sqrt{7}$

If the circles are just touching, the distance between their centres will be:
${r}_{A} + {r}_{B} = 4 + 2 \sqrt{7} \cong 9.29$

The distance from the centre of circle A to the centre of circle B = d
Using Pythagorus's theorem:
${d}^{2} = {\left(1 - 2\right)}^{2} + {\left(3 - 7\right)}^{2}$
${d}^{2} = {\left(- 1\right)}^{2} + {\left(- 4\right)}^{2}$
${d}^{2} = \left(1\right) + \left(16\right) = 17$
$d = \sqrt{17} \cong 4.12$

d is less than the sum of the two radii. Thus we can conclude that the two circles are overlapping.

They actually overlap a lot as the centre of circle A is within circle B (because ${r}_{B} > d$ )!