Two circles have the following equations $(x +8 )^2+(y -6 )^2= 64$ and $(x +4 )^2+(y -3 )^2= 144$ . 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 · 2017-12-05T12:23:41

No. Greatest possible distance between a point in one circle
and a point in other circle is $25$ unit.

Centre of first circle $(x+8)^2+(y-6)^2=8^2$ is $(-8,6)$

and radius is $8$ unit .

Centre of second circle $(x+4)^2+(y-3)^2=12^2$ is $(-4,3)$

and radius is $12$ unit . Distance between their centres is

$d=sqrt((x_1-x_2)^2+(y_1-y_2)^2)=sqrt((-8+4)^2+(6-3)^2)$ or

$d=sqrt(16+9)=5$ unit. Two circles intersect if, and only if, the

distance between their centers is between the sum $(12+8)=20$

and the difference $(12-8)=4$ of their radii. Here $4<5<12$ .

So one circle does not contain on other , they intersect at two

points. Greatest possible distance between a point in one circle

and a point in other circle is $d_g=r_1+r_2+d=8+12+5=25$

units. [Ans]

Two circles have the following equations (x +8 )^2+(y -6 )^2= 64 (x +8 )^2+(y -6 )^2= 64 and (x +4 )^2+(y -3 )^2= 144 (x +4 )^2+(y -3 )^2= 144 . 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?