Find common tangent of two circles

arze · #1 $Old$ October 20th, 2009, 11:18 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 · #2 $Old$ October 21st, 2009, 12:19 AM

Hello arze

Quote:

Originally Posted by arze $View Post$

$x^2+y^2=4x$
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 · #3 $Old$ October 21st, 2009, 06:11 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 · #4 $Old$ October 22nd, 2009, 12:13 AM

Hello arze

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

Quote:

Originally Posted by arze $View Post$

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