Find common tangent of two circles

**arze** · October 20th 2009, 09:18 PM

$x^2+y^2=4x$
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

**Grandad** · October 20th 2009, 10:19 PM

Hello arze

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.

Grandad

**arze** · October 21st 2009, 04:11 PM

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 $x^2+y^2=2x+2$

**Grandad** · October 21st 2009, 10:13 PM

Hello arze

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

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.

Grandad

Thread: Find common tangent of two circles

LinkBack

Thread Tools

Display

Find common tangent of two circles

Similar Math Help Forum Discussions

if you have two circles, how do you define the two tangents which are common to both?

Common area of circles

Common tangent?

Common external tangent

Find the points that have the common tangent line

Search Tags