Question: A point $P$ in a plane moves such that it remains at a fixed distance $r$ from a fixed point $A\equiv(r,r)$.
(i) Find the equation of the locus of point $P$ (in terms of $r$).
Another point $Q$ in the same plane moves such that it is equidistant from the point $A$ and the line $y=c$, for some constant $c\neq r$.
(ii) Find the equation of the locus of point $Q$ (in terms of $r$ and $c$).
(iii) If the locus of $P$ and that of $Q$ intersect at exactly one point, find the two possible values of $c$ (in terms of $r$).
(iv) For each of the two values of $c$ found in (iii) above, deduce the coordinates of the respective point of intersection (in terms of $r$).
Solution: (i) Let $P\equiv(x,y)$. The distance of $P$ from $(r,r)$ is $r$, so \begin{align*} \sqrt{(x-r)^2+(y-r)^2}=r. \end{align*} This gives us: \begin{align} x^2+y^2-2rx-2ry+r^2=0. \end{align} (ii)Let $Q\equiv(x,y)$. The perpendicular distance of $Q$ from $y=c$ is equal to its distance from $(r,r)$. Then, \begin{align*} y-c=\sqrt{(x-r)^2+(y-r)^2}&\Rightarrow(y-c)^2=(x-r)^2+(y-r)^2\\ &\Rightarrow -2cy+c^2=x^2-2rx-2ry+2r^2. \end{align*} Making $y$ subject of the formula: \begin{align*} y=\dfrac{x^2-2rx+(2r^2-c^2)}{2(r-c)}. \end{align*}
*In order to check my working till this point I took $r=5$ and $c=3$. This means that last equation gives the parabola $y=(x^2-10x+41)/4$ and plotting this along with the locus of $P$ gives this which seems to make sense.
(iii) I am stuck here: Looking at my graph I don't know how this is even possible.