Question

Two circles have the following equations `(x -1 )^2+(y -7 )^2= 64` and `(x +3 )^2+(y +3 )^2= 9`. 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

circles intersect

Explanation:

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 coords of the centre and r, the radius.

`rArr(x-1)^2+(y-7)^2=64" has centre (1 ,7) and r=8"`

and `(x+3)^2+(y+3)2=9" has centre (-3,-3)and r =3"`

Now compare the distance (d) between the centres with the sum of the radii.

• If sum of radii > d , then circles intersect

• If sum of radii < d , then no intersection

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 coord points"`

The 2 points here are the centres of the 2 circles.

let `(x_1,y_1)=(1,7)" and " (x_2,y_2)=(-3,-3)`

`d=sqrt((-3-1)^2+(-3-7)^2)=sqrt(16+100)≈10.77`

sum of radii = 8 + 3 = 11

Since sum of radii > d , the circles intersect.
graph{(y^2-14y+x^2-2x-14)(y^2+6y+x^2+6x+9)=0 [-40, 40, -20, 20]}
Hence 1 circle is not contained within the other and the greatest distance between a point on 1 circle and a point on the other is

d + sum of radii = 10.77 + 11 = 21.77