Two circles have the following equations ${(x - 2)}^{2} + {(y - 4)}^{2} = 36$ $(x -2 )^2+(y -4 )^2= 36$ and ${(x + 8)}^{2} + {(y + 3)}^{2} = 49$ $(x +8 )^2+(y +3 )^2= 49$ . 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

1548 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-12T14:02:17

The circles overlap and the greatest distance is $= 25.21$ $=25.21$

The center of circle $A$ $A$ is $C_{A} = (2, 4)$ $C_A=(2,4)$ and radius $r_{A} = 6$ $r_A=6$

The center of circle $B$ $B$ is $C_{B} = (- 8, - 3)$ $C_B=(-8,-3)$ and radius $r_{B} = 7$ $r_B=7$

The distance between the centers is

$C_{A} C_{B} = \sqrt{{(10)}^{2} + {(7)}^{2}} = \sqrt{149} = 12.21$ $C_AC_B=sqrt((10)^2+(7)^2)=sqrt149=12.21$

The sum of the radii is

$R = r_{A} + r_{B} = 6 + 7 = 13$ $R=r_A+r_B=6+7=13$

As,

$R > C_{A} C_{B}$ $R>C_AC_B$ , the circles overlap

The greatest distance is $= 12.21 + 6 + 7 = 25.21$ $=12.21+6+7=25.21$

graph{((x-2)^2+(y-4)^2-36)((x+8)^2+(y+3)^2-49)(y-4-7/10(x-2))=0 [-32.49, 32.46, -16.24, 16.25]}