5

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?

Sgernesto
  • 481
nightcoder
  • 125
  • 2
    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 09:04
  • Since I will have to show radius, center, etc should I not have to keep it as a symbol? – nightcoder Feb 28 '13 at 09:39
  • 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 09:41
  • @AakashM I will try to work it out that way and come back to post results. Thanks once again! – nightcoder Feb 28 '13 at 09:58
  • 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

3 Answers3

5

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.

Brian M. Scott
  • 616,228
  • 1
    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 $\sqrt{x'^2+y'^2}=\lambda\sqrt{(x'-2)^2+y'^2}$. – MichalisN Feb 28 '13 at 09:14
  • @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 09:38
  • @Michalis: Thanks a lot. I'm working with that format now to make it easier. ;) – nightcoder Feb 28 '13 at 09:40
2

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.

bubba
  • 43,483
  • 3
  • 61
  • 122
  • 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
  • 1
    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
0

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.

Macavity
  • 46,381