Two circles have the following equations (x +2 )^2+(y -5 )^2= 64 (x+2)2+(y−5)2=64 and (x +4 )^2+(y -3 )^2= 49 (x+4)2+(y−3)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?
1 Answer
circles intersect
Explanation:
If one circle is contained within the other then the following condition has to apply.
• d < R - r
d is the distance between the centres , R is the larger radius and r, the smaller.
The equation of a circle in standard form 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 centre and r, the radius.
(x+2)^2+(y-5)^2=64" has centre" (-2,5)" and " r=8
(x+4)^2+(y-3)^2=49" has centre" (-4,3)" and "r=7 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 (-2 ,5) and (-4 ,3)
d=sqrt((-4+2)^2+(3-5)^2)=sqrt8≈2.828 Now R - r = 8 - 7 = 1
here d > R - r , hence circle not contained in other.
The other possibilities are as follows.
• If sum of radii > d , then circles intersect
• If sum of radii < d , then no intersection
sum of radii = 8 + 7 = 15 > d , hence circles intersect
graph{(y^2-10y+x^2+4x-35)(y^2-6y+x^2+8x-24)=0 [-40, 40, -20, 20]}
