Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Have you got expressions for the distances $QP$ and $RP$? What happens when you set the first to $\lambda$ times the second, and simplify? (Note that $\lambda$ is just a name for a constant strictly between $0$ and $1$. Fix it at some particular value while you do some working out, if you like) – AakashM Feb 28 '13 at 9:04
Since I will have to show radius, center, etc should I not have to keep it as a symbol? – nightcoder Feb 28 '13 at 9:39
Sure, once you're happy with the process, you can do it again with a non-fixed $\lambda$. It might be easier to work through the algebraic manipulation the first time with a fixed value, that's all. – AakashM Feb 28 '13 at 9:41
@AakashM I will try to work it out that way and come back to post results. Thanks once again! – nightcoder Feb 28 '13 at 9:58
By using numbers for lambda (0.5) things turn out pretty, once I put back the lambda, there I am stuck again. – nightcoder Feb 28 '13 at 12:13
up vote 2 down vote accepted

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.

share|cite|improve this answer
Here it is: 1-$\lambda$(x^2+y^2) + x(2+4$\lambda$) + y (2+6$\lambda$) - 13 $\lambda$ + 2 I grouped by your "x2 and y2 terms with same coefficient", which was very helpful. – nightcoder Feb 28 '13 at 13:09
I think I'm getting there, I will update here. – nightcoder Feb 28 '13 at 13:11
I am surprised you don't have $\lambda^2$ terms in your equation - please double check. Assuming what you have got is correct, all that remains is to divide throughout by the coefficient of $x^2$ (or $y^2$), group the terms involving $x, x^2$ together (and $y, y^2$ together), and complete the squares. The constant term outside the squares should be positive, and you have a circle. – Macavity Feb 28 '13 at 13:29
sorry, I mistyped it. You are right, there are lots of them, since when I remove the square roots it turns the lambda into lambda squared. I got it Macavity. The radius of this beauty becomes a monster of an expression. You opened my eye with the "x^2 and y^2 with same coefficients". Thanks a lot for that. – nightcoder Feb 28 '13 at 14:29

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


Try manipulating this equation algebraically into a form that makes it clear that $C$ is a circle.

share|cite|improve this answer
If you're comfortable with it you can shift coordinates $(4,8)$ first ($Q'=(0,0),R'=(2,0),P=(x',y')$) to simplify calculations. Then you just have to solve $\sqrt{x'^2+y'^2}=\lambda\sqrt{(x'-2)^2+y'^2}$. – Michalis Feb 28 '13 at 9:14
@Brian, thanks for your hint. I actually did just that, the problem is to make the part inside the square root look like a circle algebra. Is it what the problem is all about? – nightcoder Feb 28 '13 at 9:38
@Michalis: Thanks a lot. I'm working with that format now to make it easier. ;) – nightcoder Feb 28 '13 at 9:40

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.