Review question

# If $AQ:QP=\lambda:1$, can we show $Q$ lies on a circle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9045

## Solution

A variable point $P$ on the circle $x^2+y^2=a^2$ has coordinates $(a\cos\theta,a\sin\theta)$, and $A$ is the fixed point $(c,0)$.

If $Q$ is the point on $AP$ such that $AQ:QP=\lambda:1$, find the coordinates of $Q$, and show that the locus of $Q$ is another circle. Calculate the coordinates of the centre of this circle and the length of its radius.

#### Approach 1

We begin by sketching the circle of radius $a$ about the origin, the point $P$ and the point $A$. We will consider the line $AP$ and a point $Q$ such that $AQ=\lambda QP$.

Here we have drawn the diagram in the case where $c>a$. Would our answer be any different if we had $0<c<a$ or $c<0$?

Let the co-ordinates of $Q$ be $(x,y)$ and consider the lengths we know in the following diagram. As we are given co-ordinates, we will add horizontal and vertical lines from $Q$.

By corresponding angles, the triangles with hypotenuses $AQ$ and $QP$ have the same three angles and so are similar. Hence the ratios of the opposite sides, adjacent sides and hypotenuses are equal and so

$\frac{c-x}{x-a\cos\theta} = \frac{y}{a\sin\theta-y} =\frac{\lambda}{1}.$

This rearranges to give

$c-x = \lambda x - \lambda a\cos\theta \qquad \text{and} \qquad y = \lambda a\sin\theta - \lambda y$

and hence

$Q=(x,y)=\left(\frac{\lambda a\cos\theta + c}{\lambda+1},\frac{\lambda a\sin\theta}{\lambda+1}\right).$

We want to show that this sweeps out a circle as $\theta$ varies from $0$ to $2\pi$.

We currently have equations for $x$ and $y$ in terms of $\theta$ and so, to show that this is a circle, we can start by trying to eliminate $\theta$ and expressing the locus in terms of just $x$ and $y$.

The easiest way to do this is to use the identity $\cos^2 \theta +\sin^2 \theta = 1$. Rearranging our expressions for $x$ and $y$ gives that

$\cos\theta = \left(\frac{\lambda+1}{\lambda a}\right) \left(x - \frac{c}{\lambda+1}\right) \qquad \text{and} \qquad \sin\theta = \left(\frac{\lambda+1}{\lambda a}\right) y.$

Substituting this in to our identity and dividing through by the common factor gives

$\left(x - \frac{c}{\lambda+1}\right)^2 + y^2 = \left(\frac{\lambda a}{\lambda+1}\right)^2$

and so the locus of $Q$ is a circle with centre $\left(\dfrac{c}{\lambda+1},0\right)$ and radius $\dfrac{\lambda a}{\lambda+1}$.

#### Approach 2

Consider the stripped down diagram of the problem.

We let the point $P$ have position vector $\mathbf{p}=(a\cos\theta,a\sin\theta)$, $A$ the vector $\mathbf{a}=(c,0)$ and the (currently unknown) point $Q$ the vector $\mathbf{q}$.

Travelling along the vector from $A$ to $P$ we follow the vector equation $\mathbf{r} = \mathbf{a} + \mu(\mathbf{p} - \mathbf{a}) = \left(\begin{array}{c} c \\ 0 \end{array}\right) + \mu \left(\begin{array}{c} a\cos\theta - c \\ a\sin\theta \end{array}\right),$ where $0\leq \mu \leq 1$.

It’s given that $AQ:QP=\lambda:1$, so point $Q$ is $\frac{\lambda}{\lambda+1}$ along the line from point $A$.

Therefore, if we let $\mu=\frac{\lambda}{\lambda+1}$ we will be at point $Q$ so we can then find its coordinates: $\mathbf{q}=\left(\begin{array}{c} c \\0 \end{array}\right)+\frac{\lambda}{\lambda+1}\left(\begin{array}{c} a\cos\theta - c \\ a\sin\theta \end{array}\right),$ so, using the same definition of $x$ and $y$ from before we again find that \begin{align*} x &= c+ \frac{\lambda}{\lambda+1}a\cos\theta - \frac{\lambda c}{\lambda+1}= \frac{c}{\lambda+1} + \frac{\lambda}{\lambda+1}a\cos\theta, \\ y &= \frac{\lambda}{\lambda+1}a\sin\theta \end{align*}

so again $\left(x-\frac{c}{\lambda+1}\right)^2+y^2=\frac{\lambda^2a^2}{(\lambda+1)^2},$

which leads to the same conclusion.

Show that the condition for the two circles to intersect at right angles is $\,c^2=a^2(1+2\lambda+2\lambda^2)$.

The two circles intersect at right angles if and only if there is a right-angled triangle formed by the points $O(0,0)$, $O_Q\left(\dfrac{c}{\lambda+1},0\right)$ and one of the intersection points, which we will have labelled $B$.

The triangle $OBO_Q$ is a right-angled triangle if and only if the lengths of the sides obey Pythagoras’s theorem: $a^2+\frac{\lambda^2 a^2}{(\lambda+1)^2}=\frac{c^2}{(\lambda+1)^2}$ $\iff \quad c^2=a^2(1+2\lambda+2\lambda^2)$ as required.