To find if the circles overlap we first must find the radius of each circle.
The formula for the area of a circle is:
#A = pir^2#
We can substitute for #A# and solve for #r#:
Circle A:
#15pi = pir_a^2#
#(15pi)/color(red)(pi) = (pir_a^2)/color(red)(pi)#
#(15color(red)(cancel(color(black)(pi))))/cancel(color(red)(pi)) = (color(red)(cancel(color(black)(pi)))r_a^2)/cancel(color(red)(pi))#
#15 = r_a^2#
#sqrt(15) = sqrt(r_a^2)#
#sqrt(15) = r_a#
#r_a = sqrt(15)#
Circle B:
#90pi = pir_b^2#
#(90pi)/color(red)(pi) = (pir_b^2)/color(red)(pi)#
#(90color(red)(cancel(color(black)(pi))))/cancel(color(red)(pi)) = (color(red)(cancel(color(black)(pi)))r_b^2)/cancel(color(red)(pi))#
#90 = r_b^2#
#sqrt(90) = sqrt(r_b^2)#
#sqrt(90) = r_b#
#r_b = sqrt(90)#
Now, we can graph the two circles using the equation:
#(x - color(red)(a))^2 + (y - color(red)(b))^2 = color(blue)(r)^2#
Where #(color(red)(a), color(red)(b))# is the center of the circle and #color(blue)(r)# is the radius of the circle.
Circle A:
#(x - color(red)(5))^2 + (y - color(red)(2))^2 = color(blue)(sqrt(15))^2#
#(x - color(red)(5))^2 + (y - color(red)(2))^2 = 15#
graph{((x-5)^2+(y-2)^2-15)=0 [-50. 50. -25. 25]}
Circle B:
#(x - color(red)(2))^2 + (y - color(red)(1))^2 = color(blue)(sqrt(90))^2#
#(x - color(red)(2))^2 + (y - color(red)(1))^2 = 90#
graph{((x-2)^2+(y-1)^2-90)((x-5)^2+(y-2)^2-15)=0 [-50. 50. -25. 25]}
The edges of the Circles DO NOT overlap.
However, Circle A is contained within Circle B
