How to find points of intersection of two circles algebraically?

Question

Specifically: x^2 + y^2 = 4
And: x^2 + y^2 - 4x - 4y = -4

Answer 1

x^2 + y^2 = 4
x^2 + y^2 - 4x - 4y = -4 -------------- Subtract
--------------------------------------------
4x + 4y = 8
x + y = 2
y = 2-x

x^2 + (2-x)^2 = 4
x^2 + 4 - 4x + x^2 = 4
2x^2 - 4x = 0
2x(x-2)=0
x = {0,2}
y = {2,0}

Points of intersection
{(0,2),(2,0)}

Answer 2

write x^2 = 4 - y^2. Then write the other equation as

4 - y^2 + y^2 -4(sqrt(4-y^2)) - 4y = -4

8 = 4y + 4sqrt(4-y^2)
64 = 16y^2 + 32ysqrt(4-y^2) + 16(4-y^2)
64 = 32ysqrt(4 - y^2) + 64
thus

1024y^2(4-y^2) = 0
4096y^2 = 1024y^4
or 4 = y^2
y = 2 or -2 or 0

for y = 2, x = 0 solves the first equation and the second. for y = -2, x = 0 solves the first; but trying x=0 in the second equation gives
(-2)^2 + 8 = -4 or 12 = -4, so (-2,0) is not an intersection.

and y = 0 gives x^2 - 4x = -4, or
(x-2)^2 = 0
x = 2

Thus the circles intersect at the points
(0,2) and (2,0)

Answer 3

From the first equation: (x - 3)^2 = 25 - (y - 5)^2 (x - 3)^2 = 25 - (y^2 - 10y + 25) (x - 3)^2 = 25 - y^2 + 10y - 25 (x - 3)^2 = -y^2 + 10y x - 3 = +/- sqrt(-y^2 + 10y) x = 3 +/- sqrt(-y^2 + 10y) From the second equation: x^2 = 100 - (y - 1)^2 x^2 = 100 - (y^2 - 2y + 1) x^2 = 100 - y^2 + 2y - 1 x^2 = -y^2 + 2y + 99 x = +/- sqrt(-y^2 + 2y + 99) Put those two together: 3 +/- sqrt(-y^2 + 10y) = +/- sqrt(-y^2 + 2y + 99) 3 = +/- sqrt(-y^2 + 10y) +/- sqrt(-y^2 + 2y + 99) Square both sides: 9 = (-y^2 + 10y) +/- 2*sqrt(-y^2 + 10y)*sqrt(-y^2 + 2y + 99) + (-y^2 + 2y + 99) 9 - (-y^2 + 10y) - (-y^2 + 2y + 99) = +/- 2*sqrt((-y^2 + 10y)(-y^2 + 2y + 99)) 9 + y^2 - 10y + y^2 - 2y - 99 = +/- 2*sqrt((-y^2 + 10y)(-y^2 + 2y + 99)) 2y^2 - 12y - 90 = +/- 2*sqrt(y^4 - 2y^3 - 99y^2 - 10y^3 + 20y^2 + 990y) y^2 - 6y - 45 = +/- sqrt(y^4 - 2y^3 - 99y^2 - 10y^3 + 20y^2 + 990y) Square both sides: y^4 - 6y^3 - 45y^2 - 6y^3 + 36y^2 + 270y - 45y^2 + 270y + 2025 = y^4 - 2y^3 - 99y^2 - 10y^3 + 20y^2 + 990y -45y^2 + 36y^2 + 270y - 45y^2 + 270y + 2025 + 99y^2 - 20y^2 - 990y = 0 25y^2 - 450y + 2025 = 0 y^2 - 18y + 81 = 0 (y - 9)^2 = 0 y - 9 = 0 y = 9 Substitute that back into both of the original equations: (x - 3)^2 + (9 - 5)^2 = 25 (x - 3)^2 + 4^2 = 25 (x - 3)^2 + 16 = 25 (x - 3)^2 = 25 - 16 (x - 3)^2 = 9 x - 3 = +/- sqrt(9) x - 3 = +/- 3 x = 3 +/- 3 x = 0 or 6 x^2 + (y - 1)^2 = 100 x^2 + (9 - 1)^2 = 100 x^2 + 8^2 = 100 x^2 + 64 = 100 x^2 = 100 - 64 x^2 = 36 x = +/- sqrt(36) x = +/- 6 The only value of x that satisfies both equations is 6. The only valid answer: (9,6)

Answer 4

Two Circles Intersect

Answer 5

Intersection Of Two Circles

Answer 6

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RE:
How to find points of intersection of two circles algebraically?
Specifically: x^2 + y^2 = 4
And: x^2 + y^2 - 4x - 4y = -4

Answer 7

You could multiply the first row by -1 and add it to the second, getting -4x - 4y = -8 then divide by -4 getting x+y=2

So y = 2 - x

Then plug this into the first equation: x^2 + (2-x)^2 = 4

By FOIL that's x^2 + 4 - 4x + x^2 = 4 which becomes

2x^2 - 4x = 0 which is 2x(x-2) = 0 meaning x = 0 or 2

Plug in to get y.

Answer 8

For whichever points x and y are the same for both equations, those are the points of intersection.

Answer 9

you need to solve for x and y in those 2 equations, to do this, solve for x or y in one equation, then substitute in other, solve for single variable, and plug back into equation where you solved for other variable