Circle A has a center at #(3 ,5 )# and an area of #78 pi#. Circle B has a center at #(1 ,2 )# and an area of #54 pi#. Do the circles overlap?

Answer 1

Yes

First, we need the distance between the two centres, which is #D=sqrt((Deltax)^2+(Deltay)^2)#
#D=sqrt((5-2)^2+(3-1)^2)=sqrt(3^2+2^2)=sqrt(9+4)=sqrt(13)=3.61#
Now we need the sum of radii, since: #D>(r_1+r_2);"Circles don't overlap"# #D=(r_1+r_2);"Circles just touch"# #D<(r_1+r_2);"Circles do overlap"#
#pir_1""^2=78pi# #r_1""^2=78# #r_1=sqrt78#
#pir_2""^2=54pi# #r_2""^2=54# #r_2=sqrt54#
#sqrt78+sqrt54=16.2#
#16.2>3.61#, so circles do overlap.

Proof: graph{((x-3)^2+(y-5)^2-54)((x-1)^2+(y-2)^2-78)=0 [-20.33, 19.67, -7.36, 12.64]}

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Answer 2

These overlap if #sqrt{78}+sqrt{54} ge sqrt{(3-1)^2+(5-2)^2}=sqrt{13}.#

We can skip the calculator and check #4(13)(54) ge (78-13-54)^2# or #4(13)(54) > 11^2# which it surely is, so yes, overlap.

Circle area is of course #pi r^2 # so we divide out the gratuitous #pi#s.

Our radii are squared.

#r_1^2= 78#
#r_2^2=54#

and the squared separation of the centers

#d^2=(3-1)^2+(5-2)^2=13#
Basically we want to know if #r_1+r_2 ge d #, i.e. if we can make a triangle out of two radii and the segment between the centers.

It's pretty crazy that we all automatically grab a calculator or computer and start taking square roots when the squared lengths are all nice integers.

We don't have to, but it requires a little detour. Let's use Heron's formula, call the area #Q#.
#Q= sqrt{s(s-a)(s-b)(s-c)} # where # s=(a+b+c)/2#
#Q^2 = ( (a+b+c)/2 )(( (a+b+c)/2 )-a)(( (a+b+c)/2 )-b)(( (a+b+c)/2 )-c)#
#16Q^2 = ( a+b+c )(a+b+c-2a)(a+b+c-2b)(a+b+c-2c)#
#16Q^2 = ( a+b+c )(-a+b+c)(a-b+c)(a+b-c)#

That's already superior to Heron. Nevertheless, let's move on. I'll cut out the boring parts.

#16Q^2 = 2 a^2 b^2 + 2 a^2 c^2 + 2 b^2 c^2 - ( a^4 + b^4 + c^4) #

As one might anticipate from an area formula, that is nicely symmetric. Let's make it appear less symmetrical.

# (c^2 - a^2- b^2 )^2 = a^4+b^4+c^4+2a^2b^2-2b^2c^2-2a^2c^2#

Adding,

#16Q^2 = 4 a^2 b^2 - (c^2 - a^2 - b^2)^2 #

This formula determines a triangle's squared area based on the squared lengths of its sides; if the latter are rational, the former is also.

Let's try it out. We're free to assign the sides however we like; for hand calculation its best to make #c# the largest side,
#c^2= 78#
#a^2=54#
#b^2=13#
#16Q^2 = 4 (54)(13) - (78-54-13) ^2 = 4(54)13 - 11^2 #
Even before calculating it any more, we can see we have a positive #16Q^2# so a real triangle with a positive area, so overlapping circles.
#16Q^2 =2687 #

Non-overlapping circles, or an imaginary area if we had received a negative value, indicate that the triangle is not real.

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Answer 3

To determine if the circles overlap, we need to compare the distances between their centers to the sum of their radii. If the distance between the centers is less than the sum of the radii, the circles overlap.

The distance between two points ((x_1, y_1)) and ((x_2, y_2)) is given by the formula: [ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

The radius (r) of a circle with area (A) is given by: [ r = \sqrt{\frac{A}{\pi}} ]

Using the given information, we can find the radii of Circle A and Circle B using their respective areas. Then, we can calculate the distance between their centers using the given coordinates. Finally, we compare this distance to the sum of the radii to determine if the circles overlap. If the distance is less than the sum of the radii, the circles overlap; otherwise, they do not overlap.

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