# Thread: Ellipses and Circles

1. ## Ellipses and Circles

An ellipse is drawn with maximum radius 4 cm and minimum radius 3 cm. A circle is drawn, as shown in the figure, such that any two tangents of the ellipse that meet on the circle make 90⁰ with each other. Find the radius of the circle.

2. Originally Posted by rickrishav
An ellipse is drawn with maximum radius 4 cm and minimum radius 3 cm. A circle is drawn, as shown in the figure, such that any two tangents of the ellipse that meet on the circle make 90⁰ with each other. Find the radius of the circle.
Here is an outline of one way to approach this problem. I'll work with the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$. For the final result you'll want to take a=4 and b=3.

A line through the point $(x_0,y_0)$, with slope $\lambda$, has equation $y-y_0 = \lambda(x-x_0)$. The line meets the ellipse at points where $\dfrac{x^2}{a^2}+\dfrac{(y_0 + \lambda(x-x_0))^2}{b^2} = 1.$ That equation is a quadratic in x, namely $(b^2+\lambda^2a^2)x^2 + 2\lambda a^2(y_0 -\lambda x_0)x + a^2((y-\lambda x)^2 - b^2) = 0.$ The condition for the line to be a tangent to the ellipse is that the equation for x should have equal roots, in other words "b^2 – 4ac=0". Write down that condition and simplify it, to get a quadratic equation in $\lambda$, namely $(x_0^2-a^2)\lambda^2 -2x_0y_0\lambda + y_0^2-b^2 = 0$.

That equation has two roots, say $\lambda_1$ and $\lambda_2$, which are the slopes of the two tangents to the ellipse from the point $(x_0,y_0)$. You want them to be perpendicular, in other words $\lambda_1\lambda_2 = -1$. But the product of the roots of a quadratic equation is the constant term divided by the coefficient of $\lambda^2$. That gives you the condition $\dfrac{y_0^2-b^2}{x_0^2-a^2} = -1$, which tells you that $(x_0,y_0)$ lies on the circle $x^2+y^2 = a^2+b^2$.