Two circles have the following equations $(x -1 )^2+(y -4 )^2= 36$ and $(x +5 )^2+(y -2 )^2= 49$ . 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?

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Answer 1 · 2018-04-16T13:10:02

No , they will intersect at two points.Greatest possible distance between two points of two circles is $19.68$ unit

Center of first circle $(x-1)^2+(y-4)^2=6^2$ is $(1,4)$

and radius is $6$ unit .

Center of second circle $(x+5)^2+(y-2)^2=7^2$ is $(-5,2)$

and radius is $7$ unit . Distance between their centers is

$d=sqrt((x_1-x_2)^2+(y_1-y_2)^2)=sqrt((1+5)^2+(4-2)^2)$ or

$d=sqrt 40 ~~ 6.32$ unit.

Sum of their radii is $r_1+r_2= 6+7=13$ unit

Difference of their radii is $r_2-r_1= 7-6=1$ unit

Here $(r_2-r_1) < d < (r_1+r_2) or 1<6.32<13$ , so the circles

will intersect at two points. So one circle will not contain the

other. Condition for containing one on other is $d<=|r_1-r_2|$

Greatest possible distance between two points is $2(r_1+r_2)-d$

$12+14-6.32 ~~ 19.68$ unit [Ans]

Two circles have the following equations (x -1 )^2+(y -4 )^2= 36 (x -1 )^2+(y -4 )^2= 36 and (x +5 )^2+(y -2 )^2= 49 (x +5 )^2+(y -2 )^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?