How to find the points of intersection between a circle and a parabola?

Question

Use x(squared)+2y=12 and x(squared)+y(squared)=36 as an example. I've been trying to review for my finals and this is the topic that has been very challenging for me. Thank you!

Answer 1

x² + 2y = 12

x² = 12 - 2y

x² + y² = 36

x² = 36 - y²

So, since x² = 12 - 2y and x² = 36 - y² then 12 - 2y must equal 36 - y²

12 - 2y = 36 - y²

y² - 2y - 24 = 0

(y - 6)(y + 4) = 0

y = 6 and y = -4

Since x² + 2y = 12 then, first plugging in y = 6, we get:

x² + 2(6) = 12

x² = 0

So, (0, 6) is an intersection point.

The plugging in y = -4:

x² + 2(-4) = 12

x² = 20

x = ± √20 = ±2√5

So (2√5, -4) is an intersection point and (-2√5, -4) is an intersection point.

Another method that often work nicely is Elimination:

Multiply both sides of x² + 2y = 12 by negative 1 to get: -x² - 2y = -12 then place it over the other equation and add the two equations together to eliminate "x":

-x² - 2y = -12

x² + y² = 36

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y² - 2y = 24

The solve as noted above.

Answer 2

Using your "example"

1) x² + 2y = 12

2) x² + y² = 36

Subtract 1) from 2)

3) y² - 2y = 24

Solve for y, then substitute into either of 1) or 2) to determine values of x.

Answer 3

Question:

How to find the points of intersection between a circle and a parabola?

Use x(squared)+2y=12 and x(squared)+y(squared)=36 as an example. I've been trying to review for my finals and this is the topic that has been very challenging for me. Thank you!

Solution:

Equation #1:

x(squared)+2y=12

x² + 2y = 12

Solving for x² we get:

x² = -2y + 12

Equation #2:

x(squared)+y(squared)=36

x² + y² = 36

This is a circle of radius 6 centered at the origin.

Substitute x² from Equation #1 into Equation #2 to get:

(-2y + 12) + y² = 36

y² - 2y - 24 = 0

(y + 4)(y - 6) = 0

y = -4 and 6

Plugging these y values into equation #1 gives:

x² = -2y + 12

x = ± √(12 - 2y)

y = 6:

x = ± √(12 - 2(6))

x = ± √(0) = 0

Point if intersection: (0, 6)

y = -4:

x = ± √(12 - 2(-4))

x = ± √(20)

x = ± √(2*2*5)

x = ± 2*√5

x ≅ ± 4.47

Points of intersection: (2*√5, -4) and (-2*√5, -4)

THEREFORE,

The points of intersection are:

(0, 6)

(2*√5, -4)

(-2*√5, -4)

Verified by graphing in Desmos:

y = (x² + 12) / 2

x² + y² = 36

Answer 4

Solve x² + 2y = 12 for x² This gives x² = 12 - 2y.

Substitute this into the2nd equation to get 12 - 2y + y² = 36

Solve for y by bringing everything to one side to get y² -2y - 24 = 0

Factoring this you get ( y - 6) (y + 4) = 0. Then y = 6 or -4

Plug both of these into the 1st equation and solve for x.

You will get x² = 12 - 12 = 0. so x= 0. and x² = 12 + 8 so x = √20 or 2√5

Then the points of intersection are (0, 6) and (2√5, -4)