Two circles have the following equations (x +5 )^2+(y +6 )^2= 9 (x+5)2+(y+6)2=9 and (x +2 )^2+(y -1 )^2= 1 (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?

1 Answer
CW
Sep 18, 2016

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

Explanation:

enter image source here

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<= r_Ad+rB≤rA, then Circle A contains Circle B

Given Circle A, centre (-5,-6)(−5,−6) and radius r_A=3rA=3
Circle B, centre (-2,1),(−2,1), and radius r_B=1rB=1

The first step here is to calculate d, use the distance formula :
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)d=√(x2−x1)2+(y2−y1)2

where (x_1,y_1) and (x_2,y_2)(x1,y1)and(x2,y2) are 2 coordinate points

here the two points are (-5,-6)(−5,−6) and (-2,1)(−2,1) the centres of the circles

let(x_1,y_1)=(-5,-6)(x1,y1)=(−5,−6) and (x_2,y_2)=(-2,1)(x2,y2)=(−2,1)

d=sqrt(-2-(-5)^2+(1-(-6)^2)d=√−2−(−5)2+(1−(−6)2)
=sqrt(3^2+7^2)=sqrt58=7.6158=√32+72=√58=7.6158

Sum of radii = radius of A (r_A)(rA)+ radius of B (r_B)(rB) = 3+1=4=3+1=4

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

Greatest distance = dd(the yellow segment) +r_A+r_B+rA+rB

= 7.6158+3+1=11.6158=7.6158+3+1=11.6158

Related questions
See all questions in Equation of a Circle
Impact of this question
1894 views around the world
You can reuse this answer
Creative Commons License