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.
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