# Circle A has a center at (1 ,3 ) and an area of 24 pi. Circle B has a center at (8 ,2 ) and an area of 66 pi. Do the circles overlap?

Mar 1, 2017

Yes

#### Explanation:

${A}_{1} = 24 \pi = \pi {r}_{1}^{2}$
${r}_{1}^{2} = 24$ therefore ${r}_{1} = \sqrt{24} = \sqrt{6 \cdot 4} = 2 \sqrt{6} \approx 4.9$

${A}_{2} = 66 \pi = \pi {r}_{2}^{2}$
${r}_{2}^{2} = 66$ therefore ${r}_{2} = \sqrt{66} \approx 8.12$

From $\left(1 , 3\right)$ add ${r}_{1}$ to the $x$ gives $\left(5.9 , 3\right)$

From $\left(8 , 2\right)$ subtract ${r}_{2}$ from the $x$ gives $\left(- .12 , 2\right)$

Since $- .12 < 5.9$ the circles overlap.

Graph of first circle: ${\left(x - 1\right)}^{2} + {\left(y - 3\right)}^{2} = 24$:
graph{(x-1)^2 + (y-3)^2 = 24 [-11.66, 13.65, -3.34, 9.32]}

Graph of second circle: ${\left(x - 8\right)}^{2} + {\left(y - 2\right)}^{2} = 66$:
graph{(x-8)^2 + (y-2)^2 = 66 [-15.68, 20.35, -6.93, 11.09]}