Two circles have the following equations: ${(x - 8)}^{2} + {(y - 5)}^{2} = 64$ $(x -8 )^2+(y -5 )^2= 64$ and ${(x - 7)}^{2} + {(y + 2)}^{2} = 25$ $(x -7 )^2+(y +2 )^2= 25$ . 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?

Question

1426 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-22T17:54:27

The circles overlap.
The greatest distance is $= 20.07$ $=20.07$

The center of circle A is $c_{A} = (8, 5)$ $c_A=(8,5)$ and the radius is $r_{A} = 8$ $r_A=8$

The center of circle B is $c_{b} = (7, - 2)$ $c_b=(7,-2)$ and the radius is $r_{b} = 5$ $r_b=5$

The distance between the centres of the cicles is

$d = \sqrt{{(8 - 7)}^{2} + {(5 - - 2)}^{2}}$ $d=sqrt((8-7)^2+(5--2)^2)$

$= \sqrt{1 + 49} = \sqrt{50}$ $=sqrt(1+49)=sqrt50$

The sum of the radii is $r_{A} + r_{B} = 8 + 5 = 13$ $r_A+r_B=8+5=13$

Therefore,

$d < r_{A} + r_{B}$ $d < r_A + r_B$

So, the circles overlap.

The slope of the line joining the centers is $= \frac{5 - - 2}{8 - 7} = 7$ $=(5--2)/(8-7)=7$

The greatest possible distance is $d-(max)=r_A+r_B+d$

$=8+5+sqrt50=13+7.07=20.07$

graph{( (x-8)^2) +(y-5)^2-64))((x-7)^2 + (y+2)^2-25)(y-5-7(x-8)) = 0 [-14.26, 14.24, -7.11, 7.11]}