Circle A has a center at #(2 ,12 )# and an area of #81 pi#. Circle B has a center at #(1 ,3 )# and an area of #16 pi#. Do the circles overlap? If not, what is the shortest distance between them?

Question

Circle A has a center at #(2   ,12  )# and an area of #81 pi#.  Circle B has a center at #(1  ,3   )# and an area of #16 pi#.  Do the circles overlap?  If not, what is the shortest distance between them?

Aurora Ayala · Accepted Answer

circles overlap

What we have to do here is compare the distance ( d) between the centres of the circles to the sum of the radii. • If sum of radii > d , then circles overlap • If sum of radii < d , then no overlap Before doing this, we require to find the radii of both circles. #color(orange)"Reminder"# The area (A) of a circle is #color(red)(|bar(ul(color(white)(a/a)color(black)(A=pir^2)color(white)(a/a)|)))# #color(blue)"Circle A " pir^2=81pirArrr^2=(81cancel(pi))/cancel(pi)rArrr=9# #color(blue)"Circle B " pir^2=16pirArrr^2=(16cancel(pi))/cancel(pi)rArrr=4# 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 coordinate points"# The 2 points here are (2 ,12) and (1 ,3) the centres of the circles. let # (x_1,y_1)=(2,12)" and " (x_2,y_2)=(1,3)# #d=sqrt((1-2)^2+(3-12)^2)=sqrt(1+81)=sqrt82≈9.055# sum of radii = radius of A + radius of B = 9 + 4 = 13 Since sum of radii > d , then circles overlap graph{(y^2-24y+x^2-4x+67)(y^2-6y+x^2-2x-6)=0 [-56.96, 56.94, -28.5, 28.46]}

Sadie Adams · Answer

undefined The circles do not overlap. The shortest distance between the two circles can be found by calculating the distance between their centers and then subtracting the sum of their radii.

The distance between the centers of Circle A and Circle B is given by the distance formula:

$$ \text{Distance} = \sqrt{(2 - 1)^2 + (12 - 3)^2} $$

Then, subtract the sum of the radii of Circle A and Circle B from this distance to find the shortest distance between them.