Convert to Standard Form of a Circle?

Question

I am trying to convert $$x^2 + y^2 - \frac{2(px + qy)}{p^2 + q^2 - r^2} + \frac{1}{(p^2 + q^2 - r^2)} = 0$$ and $$x^2 + y^2 - \frac{2(px + qy)}{p^2 + q^2 - r^2} + \frac{1}{((p^2 + q^2 - r^2)^2)} = 0$$

to circles of standard form $$(X-A)^2 + (Y-B)^2 = R^2$$. I've tried all sorts of algebraic manipulations, but I'm struggling! Could someone please help? Thanks :)

Answer 1

The problem being posed is confusing matters by introducing some terms that you can initially treat as opaque. For instance, let $\omega = p^2+q^2-r^2$ first, and then you have, for the first part,

$$ x^2 + y^2 - \frac{2(px+qy)}{\omega} = -\frac{1}{\omega} $$

The next key is to break the center fraction apart into its two components, and redistribute:

$$ x^2 - 2\frac{p}{\omega}x + y^2 - 2\frac{q}{\omega}y = -\frac{1}{\omega} $$

Complete the squares:

$$ x^2 - 2\frac{p}{\omega}x + \frac{p^2}{\omega^2} + y^2 - 2\frac{q}{\omega}y + \frac{q^2}{\omega^2} = -\frac{1}{\omega} + \frac{p^2}{\omega^2} + \frac{q^2}{\omega^2} $$

Factor:

$$ \left(x - \frac{p}{\omega}\right)^2 + \left(y - \frac{q}{\omega}\right)^2 = \frac{-\omega+p^2+q^2}{\omega^2} $$

Now substitute $\omega = p^2+q^2-r^2$ back in:

$$ \left(x - \frac{p}{p^2+q^2-r^2}\right)^2 + \left(y - \frac{q}{p^2+q^2-r^2}\right)^2 = \frac{r^2}{(p^2+q^2-r^2)^2} = \left(\frac{r}{p^2+q^2-r^2}\right)^2 $$

The second part can be done in essentially the same way, although it looks a little more tedious because of the squared constant term. Still, the same principles apply, I think.

Answer 2

Complete the square: $$ \left( x - \frac{p}{p^2+q^2-r^2} \right)^2 + \left( y - \frac{q}{p^2+q^2-r^2} \right)^2 = ? $$ To find the ?, expand both terms on the left and use the previous equality to work out what the term on the right should be: $$ \frac{p^2+q^2}{(p^2+q^2-r^2)^2} - \frac{1}{p^2+q^2-r^2} = \frac{p^2+q^2-p^2-q^2+r^2}{(p^2+q^2-r^2)^2} = \frac{r^2}{p^2+q^2-r^2} $$ in the first case, and $$ \frac{p^2+q^2}{(p^2+q^2-r^2)^2} - \frac{1}{(p^2+q^2-r^2)^2} = \frac{p^2+q^2-1}{(p^2+q^2-r^2)^2} $$ in the second case.

Answer 3

Ok what you can do is:

Given $(X-A)^2+(Y-B)^2=R^2$, try expanding it and see: $X^2-2XA+A^2+Y^2-2YB+B^2=R^2$

You are looking for each of the terms of this expansion which you can do by expanding also the top one, such as:

$$x^2+y^2-\frac{2(px+qy)}{p^2+q^2-r^2} + \frac{1}{p^2+q^2-r^2}=$$ $$x^2-2\frac{p}{p^2+q^2-r^2}x+y^2-2\frac{q}{p^2+q^2-r^2}y+\frac{1}{p^2+q^2-r^2}=0$$

See if you can find anything from that and maybye you'll get what you looking for.