From a point on a given circle, tangents are drawn to the ellipse. Need to find locus of chord of contact. From a point $O$ on the circle $x^2+y^2=d^2$, tangents $OP$ and $OQ$ are drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b$. Show that the locus of the midpoint of chord PQ is given by $$x^2+y^2=d^2\bigg[\frac{x^2}{a^2}+\frac{y^2}{b^2}\bigg]^2$$
I recognize that the locus of a chord whose midpoint is at $(h,k)$ is given by $\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$
I also recognize that PQ is the chord of contact, but to find its equation using the chord of contact formula I would require the coordinates of point O which I do not have.
Here I am getting the equation in terms of $x,y,h,k$, but to find the locus I need the equation entirely in the form of $h,k$, right? So how do I eliminate $x,y$ from the equation of the locus of the midpoint?

Still not what you want, but uses Joachimsthals notations.
The locus is the midpoint of the two intersection points of $$s=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$$ and (from  $s_1^2=s \cdot s_{11}$) $$(\frac{x(O)x}{a^2}+\frac{y(O)y}{b^2}-1)^2=(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1)(\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}-1)$$ which (since the square roots cancel in $\frac{x_1+x_2}{2}$ and $\frac{y_1+y_2}{2}$) is $$(x,y)=(\frac{x(O)}{\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}},\frac{y(O)}{\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}}),$$ where $x(O)^2+y(O)^2=d^2$ since $O$ is on that circle.
Writing $h=x(O), k=y(O)$ into M2
R=QQ[a,b,d]
S=R[h,k,x,y,MonomialOrder=>Eliminate 2]
I=ideal(h^2+k^2-d^2,(b^2*h^2+a^2*k^2)*x-a^2*b^2*h,(b^2*h^2+a^2*k^2)*y-a^2*b^2*k)
gens gb I

yields $$a^6b^6d^2(d^2(\frac{x^2}{a^2}+\frac{y^2}{b^2})^2-(x^2+y^2)).$$
