Find the intersection points of two circles.

Question

A circle of radius 10 at center (3,-2) and a circle of radius 10 at center (1,0) intersect. Find where they intersect.

Answer 1

circles with centers (3,-2) and (1,0)

both with radius 10 intersect at (9,6), and (-5,-8)

just graph the 2 circles on a graphing calculator and see where they intersect

or just a rough sketch will give a strong clue approximately where the intersection points are

or solve it algebraically

their equations are in general form (x-h)^2 + (y-k)^2 =r^2 where (h,k) is the center, r=radius

(x-3)^2 +(y+2)^2 = 10^2 and

(x-1)^2 +y^2 = 10^2

solve them both for y and set them equal, then solve for x

y = -2 +/-sqr(100- x^2+6x-9) for 1st equation

y =+/- sqr(100 - x^2 -2x +1) for 2nd equation

set then equal and square both sides

4+/-4sqr(91-x^2 +6x) = 101 -x^2-2x

isolate the sqr and square both sides again

16(91-x^2 +6x) = (-x^2-2x +95)^2 = (x^2 +2x-95)^2 = x^4+2x^3 -95x^2 +2x^3+4x^2 -95x -95x^2 -110x + 95^2

-16x^2 +96x + 16(91) = x^4 +4x^3 -106x^2-205x+9025

x^4 +4x^3- 90x^2 -301x+7549=0

there's a "small" arithmetic error in the above tedious calculations somewhere,

but if you graph this 4th degree equation you get close to 8 abd -5 as the x intercepts

along with 2 extraneous solutions due to squaring, multiplying by a variable

another check on (9,6) and (-5,-8) as the solutions is

set the radius squareds equal

100=100

x^2-6x +9 + y^2+4y+4 =x^2 -2x +1 +y^2, the squared terms all cancel out leavig

y =x-3

plug in the two intersection points

6=9-3

and

-8=-5-3

they both work

Answer 2

We've told you that these circles intersect. If you did not know

this, you should determine whether they intersect first.

The general equation of a circle is:

r^2 = (x-x_o)^2 + (y-y_o)^2

where:

r is the radius

x_o is the x-coordinate of the origin

y_o is the y-coordinate of the origin

So the first circle has equation:

100 = (x-3)^2 + (y+2)^2

The second:

100 = (x-1)^2 + y^2

The easiest way to solve this problem is to subtract one equation from the other:

100=(x-3)^2 + (y+2)^2

-

100=(x-1)^2 + y^2

=

0=(x-3)^2 +(y+2)^2 - [(x-1)^2 +y^2]

0=(x-3)^2 +(y+2)^2 - (x-1)^2 -y^2

0=x^2-6x+9 + y^2+4y+4 - (x^2-2x+1) -y^2

0=x^2-6x+9 + y^2+4y+4 -x^2+2x-1 -y^2

0=-4x+12+4y

0=-x+3+y

y=x-3

We substitute into one of the two equations:

100=(x-1)^2 + y^2

100=(x-1)^2 + (x-3)^2

100=x^2-2x+1 + x^2-6x+9

100=2x^2-8x+10

50=x^2-4x+5

0=x^2-4x-45

0=(x-9)(x+5)

x = 9,-5

y=x-3

y=9-3

y=6

(9,6)

y=x-3

y=-5-3

y=-8

(-5,-8)

Check that the points we found are on the circles:

First point: (9,6)

100=(9-3)^2 + (6+2)^2

100=6^2 + 8^2

100=36+64

100=100

100=(x-1)^2 + y^2

100=(9-1)^2 + 6^2

100=64+36

100=100

Second point: (-5,-8)

100=(-5-3)^2 +(-8+2)^2

100=(-8)^2 +(-6)^2

100=64+36

100=100

100=(x-1)^2 +y^2

100=(-5-1)^2 +(-8)^2

100=36+64

10=100

The circles intersect at (9,6) and (-5,-8).