$$
\begin{align}
QX^2&=XM(XM+PM)\tag1\\
\frac{QX+Q'Q}{P'P}&=\frac{XM}{PM}\tag2\\
QX^2+PQ^2&=(PM+XM)^2\tag3\\
PQ^2&=PM(XM+PM)\tag4\\
\frac{QX^2}{PQ^2}&=\frac{XM}{PM}\tag5\\
\frac{QX^2}{PQ^2}&=\frac{QX+Q'Q}{P'P}\tag6\\
0&=P'P(QX)^2-PQ^2(QX)-PQ^2Q'Q\tag7
\end{align}
$$
Explanation:
$(1)$: power of the point $X$
$(2)$: $\triangle PMP'\cong\triangle XMQ'$
$(3)$: Pythagorean Theorem
$(4)$: subtract $(1)$ from $(3)$
$(5)$: divide $(1)$ by $(4)$
$(6)$: $(2)$ and $(5)$ are equal
$(7)$: write as a polynomial in $QX$
