Question

Two circles have the following equations `(x +3 )^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?

Answer 1

One circle contains the other.

Explanation:

What we have to do here is compare the distance (d ) between the centres of the circles with the sum or difference of the radii.
There are 3 possible outcomes.

• If sum of radii > d , then circles intersect

• If sum of radii < d , then no intersection

• If difference of radii > d , then 1 circle contains the other

The standard form of the equation of a circle is.

`color(red)(|bar(ul(color(white)(a/a)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(a/a)|)))`
where (a ,b) are the coordinates of the centre and r, the radius.

`(x+3)^2+(y-6)^2=64rArr" centre"=(-3,6),r=8`

`(x+4)^2+(y-3)^2=144rArr" centre"=(-4,3),r=12`

To calculate d, use the `color(blue)"distance formula"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(a/a)|))`
where `(x_1,y_1)" and " (x_2,y_2)" are 2 points"`

The 2 points here are (-3 ,6) and (-4 ,3) the centres of the circles.

`d=sqrt((-4+3)^2+(3-6)^2)=sqrt(1+9)≈3.162`

difference of radii = 12(larger)- 8(smaller) = 4

Since difference of radii > d, then 1 circle is contained in the other.
graph{(y^2-12y+x^2+6x-19)(y^2-6y+x^2+8x-119)=0 [-40, 40, -20, 20]}