Question: A circle is drawn with its center at (8,0) and with radius r such that the circle cuts the ellipse $x^2 + 4y^2 = 16$ at right angles. Find the radius of the circle.
I understand that the tangent of a circle at the point they intersect is the normal of the ellipse. I don't where to go from there to get the radius. Thank you
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Hint 1 The straight line from $(8, 0)$ to the intersection point must be tangent to the ellipse. Using $y^\prime = 2 y$, this property remains true in the $(x, y^\prime)$ plane, where the ellipse is now a circle.
Hint 2 The ellipse can be parametrized by $(x, y) = (4 \cos(t), 2 \sin(t))$. Find the values of $t$ where the tangent vector is proportional to $(x-8, y)$.
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The equation of the circle is $$(x-8)^2+y^2=R^2\leftrightarrow x^2+y^2-16x=R^2-64.$$ The equations of the tangent lines of these two conics at an intersection point $(x_0,y_0)$ are respectively (see here) $$x_0x+4y_0 y=16\quad\mbox{and}\quad x_0x+y_0 y-8(x+x_0)=R^2-64.$$ These two lines are orthogonal iff the following scalar product is zero $$0=(x_0,4y_0)\cdot (x_0-8,y_0)=x_0^2+4y_0^2-8x_0=16-8x_0\implies x_0=2.$$ Hence $y_0^2=(16-x_0^2)/4=3$ and $$R=\sqrt{(x_0-8)^2+y_0^2}=\sqrt{39}.$$
The circle and ellipse equations: $$(x-8)^2+y^2=r^2; \frac{x^2}{4^2}+\frac{y^2}{2^2}=1.$$ The slopes of their tangents at the common intersection point $x_0$: $$2(x-8)+2y\cdot y'=0 \Rightarrow y'(x_0)=-\frac{x_0-8}{y(x_0)},$$ $$\frac{2x}{4^2}+\frac{2yy'}{2^2}=0 \Rightarrow y'(x_0)=-\frac{x_0}{4y(x_0)}.$$ For the tangent lines to be perpendicular the product of slopes must be $-1$: $$(x_0-8)x_0=-4y^2(x_0) \Rightarrow x_0=2.$$ Now from the ellipse equation we will find $y_0^2$: $$y_0^2=3.$$ Finally from the circle equation we will find the radius: $$r^2=(2-8)^2+3=39 \Rightarrow r=\sqrt{39}.$$