Two circles have the following equations ${(x + 3)}^{2} + {(y - 6)}^{2} = 64$ $(x +3 )^2+(y -6 )^2= 64$ and ${(x + 7)}^{2} + {(y - 3)}^{2} = 4$ $(x +7 )^2+(y -3 )^2= 4$ . 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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Two circles have the following equations ${(x + 3)}^{2} + {(y - 6)}^{2} = 64$ $(x +3 )^2+(y -6 )^2= 64$ and ${(x + 7)}^{2} + {(y - 3)}^{2} = 4$ $(x +7 )^2+(y -3 )^2= 4$ . 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 · 2016-08-29T08:59:30

One circle contains the other

Explanation:

The distance between the centres (-2,6) and(-7 ,3) is 5 ..use Pythagoras

Radius of the first circle is 8 and of the second 2.
Draw a line through the centres and mark in where the radii cross the line to prove to yourself that one circle contains the other.

This all rests on our knowing the general equation of a circle
${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$ gives the centre(a,b) and radius r

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Two circles have the following equations (x+3)2+(y−6)2=64(x +3 )^2+(y -6 )^2= 64 and (x+7)2+(y−3)2=4(x +7 )^2+(y -3 )^2= 4 . 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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Explanation:

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