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Math Help - Ellipses and Circles

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    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.
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    Quote Originally Posted by rickrishav View Post
    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.
    Last edited by Opalg; December 9th 2010 at 01:25 AM.
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