Circles are drawn passing through the origin O to intersect the coordinate axes at points P and Q such that $(m)(OP)+(n)(OQ)=k$, then the fixed point satisfying all such circles is?
(A) $\left(m,n\right)$
(B) $\left(\dfrac{m^2}{k},~\dfrac{n^2}{k}\right)$
(C) $\left(\dfrac{mk}{m^2+n^2},~\dfrac{nk}{m^2+n^2}\right)$
(D) $\left(k,k\right)$
I can't make any sense of what I tried. The answer given is option (C).
I took the equation of one of these circles as $x^2+y^2+2gx+2fy=0$, and I drew a diagram and concluded that $OP=-2g$ and $OQ=-2f$. I put that in the given equation in $OP$ and $OQ$. Then I tried putting the coordinates $\left(m,n\right)$ in the equation of the circle, because I could then substitute $k$, but I couldn't make sense of what I was doing. Can someone please tell me how to solve this?
Since the circle always passes through the origin, let its equation be :$$x^2+y^2=2gx+2fy$$ This circle makes intercepts $2g$ and $2f$ on the axes and its center is at $(g,f)$. Therefore if we consider signs of the intercepts, we have $OP=2g$ and $OQ=2f$ .
The other given condition is this:$$2gm+2nf=k$$ Consider the fixed line $2mx+2ny=k$ . Now its clear that the above condition simply states that the center of the circle $C$ will lie on this line. This was the first tricky part of this question.
Now we know that we have a family of circles that pass through the origin and have the centers on a fixed line, but how to find the common point of this family? This is the second tricky part of the question.
To tackle this, you have to think geometrically. Choose an arbitrary point $C$ on the fixed line as the center of the circle. The radius of this circle is simply the distance $\overline{OC}$. Say the reflection of the origin $O$ about the fixed line is the point $O\space '$. By symmetry we have $\overline{OC}=\overline{O\space'C}$ and hence the point $O\space'$ will always lie on the circle. Following diagram illustrates this:
Now all you have to do is find the coordinates of the point O' .
EDIT: I just learned how to make GIFs on linux, so I made one that nicely shows that the above family of circles passes through a fixed point :