Find common tangent of two circles

• Oct 20th 2009, 09:18 PM
arze
Find common tangent of two circles
$x^2+y^2=4x$
The tangents at the points P and Q on this circle touch the circle $x^2+y^2=1$ at the points R and S. Find the coordinates of the point of intersection of these tangents, and obtain the equation of the circle through the points P, Q, R, and S.

I need to know how to find the equation of the common tangent, but I don't know where to start.
Thanks
• Oct 20th 2009, 10:19 PM
Hello arze
Quote:

Originally Posted by arze
$x^2+y^2=4x$
The tangents at the points P and Q on this circle touch the circle $x^2+y^2=1$ at the points R and S. Find the coordinates of the point of intersection of these tangents, and obtain the equation of the circle through the points P, Q, R, and S.

I need to know how to find the equation of the common tangent, but I don't know where to start.
Thanks

By completing the square, re-write $x^2+y^2=4x$ as $(x-2)^2+y^2=4$ to note that this circle has centre $(2,0)$ and radius $2$. The circle $x^2+y^2=1$ has centre $(0,0)$ and radius $1$. So the circles are positioned as in the attached diagram.

Using similar triangles, can you see that $TA = 2TO$?

The coordinates of $T$ are therefore fairly obvious.

The centre of the circle through $P,Q,R, S$ lies on the perpendicular bisector of $SP$ - and also, by symmetry, on the $x$-axis. Can you work out the coordinates of this centre, and the radius of the circle now? Its equation is then very straightforward to write down.

Get back to us if you still need more help.

• Oct 21st 2009, 04:11 PM
arze
Thanks! The gradient of the tangent through P and S would be 1/2, right? so with th point (-2,0) we find the equation of the tangent?
$y=\frac{1}{2}(x+2)$ then substitute into $x^2+y^2=4x$ and $x^2+y^2=1$ for the x-values of P and S
$P(2,2)$ and $S(-\frac{4}{5},\frac{3}{5})$
Mid( $\frac{3}{5},\frac{13}{10}$)
So $y-\frac{13}{10}=-2(x-\frac{3}{5})$
$y+2x=\frac{5}{2}$
when y=0 $x=\frac{5}{4}$
radius = $\sqrt{(\frac{5}{4}-2)^2+2^2}=\sqrt{\frac{73}{16}}$
$(x-\frac{5}{4})^2+y^2=\frac{73}{16}$
$x^2+y^2-\frac{5}{2}x-3=0$
But the answer is supposed to be $x^2+y^2=2x+2$
• Oct 21st 2009, 10:13 PM
Hello arze

You've got a mistake here at the start which of course puts everything else out.
Quote:

Originally Posted by arze
Thanks! The gradient of the tangent through P and S would be 1/2, right?

No. Note that $\sin \angle ATP =\frac{AP}{AT}=\frac12$

$\Rightarrow \angle ATP = 30^o$

So the gradient of $TP = \tan 30^o=\frac{1}{\sqrt3}$

However, rather than find $P,Q,R,S$ by coordinate geometry, it's much easier to use trigonometry - especially since we now know that $\angle ATP = 30^o$. Here's a start:

$S$ has $x$-coordinate $=-OS\cos\angle SOT= -OS \cos 60^o = -\tfrac12$, and $y$-coordinate = $OS \sin 60^o=\frac{\sqrt3}{2}$

You should be able to continue from here.