# Two circles have the following equations (x +5 )^2+(y +6 )^2= 9  and (x +2 )^2+(y -1 )^2= 1 . Does one circle contain the other? If not, what is the greatest possible distance between a point on one circle and another point on the other?

Sep 18, 2016

One circle does not contain the other. Greatest distance $= 11.6158 .$

#### Explanation: Compare the distance (d) between the centres of the circles to the sum of the radii.

1) If the sum of the radii $>$d, the circles overlap.
2) If the sum of the radii $<$d, then no overlap.
3) If $d + {r}_{B} \le {r}_{A}$, then Circle A contains Circle B

Given Circle A, centre $\left(- 5 , - 6\right)$ and radius ${r}_{A} = 3$
Circle B, centre $\left(- 2 , 1\right) ,$ and radius ${r}_{B} = 1$

The first step here is to calculate d, use the distance formula :
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

where $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$ are 2 coordinate points

here the two points are $\left(- 5 , - 6\right)$ and $\left(- 2 , 1\right)$ the centres of the circles

let$\left({x}_{1} , {y}_{1}\right) = \left(- 5 , - 6\right)$ and $\left({x}_{2} , {y}_{2}\right) = \left(- 2 , 1\right)$

d=sqrt(-2-(-5)^2+(1-(-6)^2)
$= \sqrt{{3}^{2} + {7}^{2}} = \sqrt{58} = 7.6158$

Sum of radii = radius of A $\left({r}_{A}\right)$+ radius of B $\left({r}_{B}\right)$ $= 3 + 1 = 4$

Since sum of radius $<$d, then no overlap of the circles
no overlap => no containment

Greatest distance = $d$(the yellow segment) $+ {r}_{A} + {r}_{B}$

$= 7.6158 + 3 + 1 = 11.6158$