+45 An ellipse is defined by the equation x^2 /9 + y^2 /4 = 1. Compute the radius of the largest circle that is internally tangent to the ellipse at (3, 0) and intersects the ellipse only at (3, 0).
+2401 Let r be the radius of the circle.
The circle must have a center of (3 - r, 0).
So our formula is (x - 3 + r)^2 + y^2 = r^2
9(x - 3 + r)^2 + 9y^2 = 9r^2
x^2 /9 + y^2 /4 = 1
4x^2 + 9y^2 = 36
Subtracting the two formulas.
9(x - 3 + r)^2 - 4x^2 = 9r^2 - 36
5x^2 + (18r - 54)x + (117 - 54r) = 0
There is one solution when the discriminant is 0.
(18 - 54r)^2 - 4(5)(117 - 54r) = 0
r = 2(2 + sqrt(130))/27 (note that r can't be negative so it's not the other option)
Hope this was helpful.
=^._.^=
+118608 Thanks Catmg,
That looks like a bit of a challenge. :)
I just thought I'd take a graphical look at your result. Looks reasonable :)
https://www.desmos.com/calculator/zbydksn1f1
