See attachment, thanks for looking
The ellipse will have foci at (-d, 0) and (d, 0). Let the sum of the distances $\displaystyle F_1P + F_2P = c$. Then the equation of the ellipse is $\displaystyle \sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$. I simplified this equation to $\displaystyle \frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$. Now can you show that the circle $\displaystyle x^2 + y^2 = \frac{c^2}{4}$ has the desired property?
Ok, here goes:
$\displaystyle \sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$
$\displaystyle \sqrt{y^2 + (x+d)^2} = c - \sqrt{y^2 + (x-d)^2}$
$\displaystyle y^2 + (x+d)^2 = c^2 - 2c\sqrt{y^2 + (x-d)^2} + y^2 + (x-d)^2$
$\displaystyle (x+d)^2 - (x-d)^2 - c^2 = -2c\sqrt{y^2 + (x-d)^2}$
$\displaystyle x^2 + 2xd + d^2 - (x^2 - 2xd + d^2) - c^2 = -2c\sqrt
{y^2 + (x-d)^2}$
$\displaystyle x^2 + 2xd + d^2 - x^2 + 2xd - d^2 - c^2 = -2c\sqrt{y^2 + (x-d)^2}$
$\displaystyle 4xd - c^2 = -2c\sqrt{y^2 + (x-d)^2}$
$\displaystyle (4xd - c^2)^2 = 4c^2(y^2 + (x-d)^2)$
$\displaystyle 16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2(x-d)^2$
$\displaystyle 16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2(x^2 - 2xd + d^2)$
$\displaystyle 16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2x^2 - 8xdc^2 + 4c^2d^2$
$\displaystyle 16x^2d^2 + c^4 = 4c^2y^2 + 4c^2x^2 + 4c^2d^2$
$\displaystyle (16d^2 - 4c^2)x^2 - 4c^2y^2 = 4c^2d^2 - c^4$
$\displaystyle 4(4d^2 - c^2)x^2 - 4c^2y^2 = c^2(4d^2 - c^2)$
$\displaystyle \frac{4x^2}{c^2} - \frac{4y^2}{4d^2 - c^2} = 1$
$\displaystyle \frac{4x^2}{c^2} + \frac{4y^2}{c^2 - 4d^2} = 1$