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I was looking for a quatric equation that combines the equation of the circle and ellipse, which are:

\begin{align*} Circle: (x - m)^2 + (y - n)^2 &= r^2 \\\ Ellipse: \frac{(x - o)^2}{a^2} + \frac{(y - p)^2}{b^2} &= 1 \end{align*}

I found one that works for what I need in this post, which is:

\begin{align*} a^4 \left(b^2-(p-y)^2\right)^2+2 a^2 b^2 \left((p-y)^2-b^2\right) \left(m^2-2 m o-n^2+2 n y+o^2+r^2-y^2\right)+b^4 \left(m^2-2 m o+n^2-2 n y+o^2-r^2+y^2\right)^2=0 \end{align*}

I want to know the procedure to get this result and how to get the equation for x.

Erika Pan
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1 Answers1

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Given that

\begin{align*} Circle: (x - m)^2 + (y - n)^2 &= r^2 \\\ Ellipse: \frac{(x - o)^2}{a^2} + \frac{(y - p)^2}{b^2} &= 1 \end{align*}

they can also be written as

\begin{align*} Circle: (x - m)^2 + (y - n)^2-r^2 &=0 \\\ Ellipse: \frac{(x - o)^2}{a^2} + \frac{(y - p)^2}{b^2}-1 &= 0 \end{align*}

now multiplying both $L.H.S$ and equating result to $zero$ will give you combined equation of given circle and ellipse,I think you can take it from there by multiplying and expanding each term..

Edit:Clarification to OP's Further doubt:

In order to find the quartic in terms of $x$ only,it would be tedious

let $x=m+rcos(z)$,$y=n+rsin(z)$ now

$(y-p)^2=(n-p+rsin(z))^2$

$(y-p)^2=(n-p)^2+2(n-p)(rsin(z))+r^2sin^2(z)$

$(y-p)^2-r^2sin^2(z)-(n-p)^2=2(n-p)(rsin(z))$

squaring both sides,

$((y-p)^2-r^2sin^2(z)-(n-p)^2)^2=4(n-p)^2(r^2sin^2(z)).....(1)$

now consider equation of ellipse

$\frac{(x - o)^2}{a^2} + \frac{(y - p)^2}{b^2}-1=0$

cross multiplying terms and rearranging $(y-p)^2=b^2-\frac{b^2}{a^2}(x-o)^2$

subtracting $r^2sin^2(z)+(n-p)^2$ from both sides

$(y-p)^2-r^2sin^2(z)-(n-p)^2=b^2-\frac{b^2}{a^2}(x-o)^2-r^2sin^2(z)-(n-p)^2$

squaring both sides,

$((y-p)^2-r^2sin^2(z)-(n-p)^2)^2=(b^2-\frac{b^2}{a^2}(x-o)^2-r^2sin^2(z)-(n-p)^2)^2....(2)$

substituting value of $((y-p)^2-r^2sin^2(z)-(n-p)^2)^2$ from equation (1) into (2)

$4(n-p)^2(r^2sin^2(z))=(b^2-\frac{b^2}{a^2}(x-o)^2-r^2sin^2(z)-(n-p)^2)^2...(3)$

now substitute $sin^2(z)=1-cos^2(z)$ in $(3)$

$4(n-p)^2(r^2(1-cos^2(z)))=(b^2-\frac{b^2}{a^2}(x-o)^2-r^2(1-cos^2(z))-(n-p)^2)^2...(4)$

now substitute $cos^2(z)=(\frac{x-m}{r})^2$ in $(4)$

$4(n-p)^2(r^2(1-(\frac{x-m}{r})^2))=(b^2-\frac{b^2}{a^2}(x-o)^2-r^2(1-(\frac{x-m}{r})^2)-(n-p)^2)^2...(5)$

now you can take expand it and rearrange to get desired quartic you want

Dheeraj Gujrathi
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