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

## Suggestion

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.

We can experiment with this GeoGebra applet for this problem.

As we move the point $P$ around its circle, what happens to $Q$?

Try with $A$ in different positions along the axis. What if $A$ is outside the circle?

You can change the value of $\lambda$ by sliding $Q$ along the line.

The point $Q$ lies some way along the line $AP$. Can you find the equation of this line?

How could we use the fact $AQ:QP=\lambda:1$? Could we find the lengths of the two line segments $AQ$ and $QP$?

Alternatively, is there a way of using $AQ:QP=\lambda:1$ without knowing the lengths of the line segments?

Could we find the coordinates of $Q$ as a function of $\theta$? ($\theta$ is defined in the question.)

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

For the second part, this GeoGebra tool might help.

We can now drag the point $O_Q$, the centre of the second circle (the point $A$ is not shown here).

When the two circles intersect at right angles, what can we say about the triangle whose vertices are the centres of the two circles and one of their points of intersection?