Prove it is a circle So I have this question:

Let $Q = (4, 8)$, $R = (6, 8)$ and $P = (a, b)$. Let $\lambda\in\mathbb R$ with $0 < \lambda < 1$.
Consider $C =\{P: |QP| = \lambda|RP|\}$ 
Give an equation to $C$ and prove its a circle.

I'm trying to figure out how to interpret the $\lambda$ symbol to come up with the an expression for $C$, which I have to prove is a circle.
I did work out the distances $PQ$ and $QR$, the $\lambda$ symbol is just puzzling me.
I tried to fix $\lambda$ and divide the two distance equations, but it leads me nowhere.
Can anyone give me some directions?
 A: HINT: $\lambda$ is just some constant between $0$ and $1$. Consider the point $P=\langle x,y\rangle$: $$|QP|=\sqrt{(x-4)^2+(y-8)^2}\;,$$ and $$|RP|=\sqrt{(x-6)^2+(y-8)^2}\;,$$ so $C$ is the set of all points $P=\langle x,y\rangle$ such that
$$\sqrt{(x-4)^2+(y-8)^2}=\lambda\sqrt{(x-6)^2+(y-8)^2}\;.$$
Try manipulating this equation algebraically into a form that makes it clear that $C$ is a circle.
A: Maybe the problem is that you don't know what sorts of equations represent circles, so, as you are doing the algebraic manipulations, you have no target/destination in mind.
Any equation of the form
$$a(x^2 + y^2) + bx + cy +d = 0$$
is a circle. If you don't know why, please ask. The key characteristics are that the $x^2$ and $y^2$ terms have the same coefficient, and there is no $xy$ term.
So, take the equation in Brian Scott's answer, and see if you can massage it into this form. If you can do that, then you will know you have a circle.
After you have the equation in the form above, it's easy to show that the center of the circle is at the point $(-\frac{b}{2a}, -\frac{c}{2a} )$. You can see this by "completing the squares" as Macavity said.
A: Using vectors may simplify the algebra.  For instance, with $P, Q, R$ as position vectors, we have the locus of points in set $C$ to be:  
$|P- Q|^2 = \lambda^2 |P-R|^2$.  
Using dot products, expanding and simplifying, one gets:
$\big|P - \dfrac{Q - \mu R}{1-\mu}\big|^2 = \mu \dfrac {|Q-R|^2}{(1-\mu)^2} $
where $\mu = \lambda^2$
from which it is easy to recognise the circle form.
