Two circles of radius r are externally tangent to each other and internally tangent to the ellipse x^2 + 2y^2 = 6 as shown below. Find r.
The centers of the circles will be (r, 0) and (-r, 0)
The equations of the circles are
(x -r)^2 + y^2 = r^2 ⇒ x^2 - 2xr + r^2 + y^2 = r^2 ⇒ y^2 = 2xr - x^2
(x + r)^2 + y^2 = r^2 ⇒ x^2 + 2xr + r^2 + y^2 = r^2 ⇒ y^2 = -2xr - x^2 = -(2xr + x^2)
Subbing in y^2 in the equation of the ellipse we have this system
x^2 + 2(2xr - x^2) = 6
x^2 - 2 (2xr + x^2) = 6
x^2 + 4xr - 2x^2 = 6
x^2 - 4xr + 2x^2 = 6 add these
2x^2 = 12
x^2 = 6
x = sqrt (6)
Using x^2 + 4xr - 2x^2 = 6 to solve for r
(sqrt (6))^2 + 4 (sqrt 6) r - 2(sqrt 6)^2 = 6
6 + 4sqrt(6) r - 12 = 6
4sqrt (6) r = 12
sqrt (6)r = 3
r = 3 / sqrt (6) = 3/ ( sqrt 3 * sqrt 2) = sqrt (3/2)
The graph below shows what we have.....the circles are tangent to each other and internally tangent to the ellipse at the endpoints of the major axis of the ellipse......