# Thread: circle belonging to a ellipse

1. ## circle belonging to a ellipse

See attachment, thanks for looking

2. The ellipse will have foci at (-d, 0) and (d, 0). Let the sum of the distances $F_1P + F_2P = c$. Then the equation of the ellipse is $\sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$. I simplified this equation to $\frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$. Now can you show that the circle $x^2 + y^2 = \frac{c^2}{4}$ has the desired property?

3. Originally Posted by icemanfan
The ellipse will have foci at (-d, 0) and (d, 0). Let the sum of the distances $F_1P + F_2P = c$. Then the equation of the ellipse is $\sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$. I simplified this equation to $\frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$. Now can you show that the circle $x^2 + y^2 = \frac{c^2}{4}$ has the desired property?
How did u simplify your the equation to the following $\frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$.??? I tired to square both sides but it just leaves one big mess and i can never get down to the same equation you got.

4. Ok, here goes:

$\sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$

$\sqrt{y^2 + (x+d)^2} = c - \sqrt{y^2 + (x-d)^2}$

$y^2 + (x+d)^2 = c^2 - 2c\sqrt{y^2 + (x-d)^2} + y^2 + (x-d)^2$

$(x+d)^2 - (x-d)^2 - c^2 = -2c\sqrt{y^2 + (x-d)^2}$

$x^2 + 2xd + d^2 - (x^2 - 2xd + d^2) - c^2 = -2c\sqrt
{y^2 + (x-d)^2}$

$x^2 + 2xd + d^2 - x^2 + 2xd - d^2 - c^2 = -2c\sqrt{y^2 + (x-d)^2}$

$4xd - c^2 = -2c\sqrt{y^2 + (x-d)^2}$

$(4xd - c^2)^2 = 4c^2(y^2 + (x-d)^2)$

$16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2(x-d)^2$

$16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2(x^2 - 2xd + d^2)$

$16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2x^2 - 8xdc^2 + 4c^2d^2$

$16x^2d^2 + c^4 = 4c^2y^2 + 4c^2x^2 + 4c^2d^2$

$(16d^2 - 4c^2)x^2 - 4c^2y^2 = 4c^2d^2 - c^4$

$4(4d^2 - c^2)x^2 - 4c^2y^2 = c^2(4d^2 - c^2)$

$\frac{4x^2}{c^2} - \frac{4y^2}{4d^2 - c^2} = 1$

$\frac{4x^2}{c^2} + \frac{4y^2}{c^2 - 4d^2} = 1$

5. i know i need to confirm that substituting into the ellipse and into the circle will result in the same values of x

this k is equal to . How does one know that k is equal to this??

6. Also, have another question. How does one know that the center of the circle in this problem is at c^2/4??