I spent some time on this question earlier today - the most difficult part was understanding the language used in the question and being able to visualise the point stated in the question as a point in the plane. After some wrestling I have managed to get an answer similar to the text but I appear to out by a factor of two, i.e. my solution is half of that quoted in the text. I'll stop waffling and give you the question, apologies for my lack of a diagram and if my algebra isn't perfect.
Question is as follows:
The equations of the sides of a rectangle are $x=0$, $x=2a$, $y=0$ and $y=a$. A point moves so that the sum of the squares of its distances from the sides of the rectangle is $k$, a constant. Show that the path of the point is a circle and find the coordinates of its centre.
If the radius of the circle is $5a$, find $k$.
Solution gives: $(a,\frac{a}{2})$ and $k=105a^2$
My working is as follows and apologies once again for lack of a diagram.
Let $P(\alpha,\beta)$ be a point on the locus.
Then we have: $$(\alpha-2a)^2+\alpha^2+(\beta-a)^2+\beta^2=k$$ $$\implies 2\alpha^2+2\beta^2-4a\alpha-2a\beta+4a^2+a^2-k=0$$ $$\implies\alpha^2+\beta^2-2a\alpha-a\beta+\frac{5a^2}{2}-\frac{k}{2}=0\qquad(1)$$
This has the general form of a circle i.e., $\alpha^2+\beta^2+2f\alpha+2g\beta+c=0$ with centre $(-f,-g)$ and radius, $$r=\sqrt{f^2+g^2-c}\qquad(2)$$
Therefore, as $x=\alpha,\text{and},\ y=\beta$, equation of circle is: $$x^2+y^2-2ax-ay+\frac{5a^2}{2}-\frac{k}{2}=0$$ $$\therefore\quad\text{locus is a circle with centre at}(a,\frac{a}{2})$$ $$\text {and radius},r=\sqrt{a^2+(\frac{a}{2})^2-\frac{5a^2}{2}+\frac{k}{2}}$$
Now, when $r=5a$, we have: $$25a^2=\frac{5a^2}{4}-\frac{5a^2}{2}+\frac{k}{2}$$ $$\implies100a^2=-5a^2+2k$$ $$\therefore\qquad k=\frac{105a^2}{2}$$
I hope the above is clear and there are no errors in the algebra. If anyone can offer any advice with regards to whether or not my solution is correct then I would be very grateful.
Many thanks in advance.