Tangents of circles

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Tangents of circles

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Source: BrMO R1 2006/07 P4

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#1 h Jun 2, 2014, 5:57 AM • 2 Y

Y by Adventure10, Mango247

Two touching circles $S$ and $T$ share a common tangents which meets $S$ at $A$ and $T$ at $B$ . Let $AP$ be a diameter of $S$ and let the tangent from $P$ to $T$ touch it at $Q$ . Show that $AP=PQ$ .

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#2 h Jun 2, 2014, 10:02 AM • 2 Y

Y by Adventure10, Mango247

Let the common tangency point be $X$ . Let the centre of $S$ be $U$ and centre of $T$ be $V$ .
As $UA || VB$ So by angle chasing we see that $\angle AXB = 90^0$ So when $BX$ is produced to meet at $S$ at a point $R$ So $\angle AXR = 90^0$ (as $\angle AXB = 90^0$ ). So $AR$ is a diameter
So $R=P$ .
and $X,P,B$ are collinear.
Also by angle chasing we see that $\angle XBA =\angle XAP \Longrightarrow AP$ is tangent to $\odot AXB$ $\Longrightarrow AP^2 = PX.PB$
So $PQ$ is tangent to $T$ $\Longrightarrow PQ^2 = PX.PB$
So $AP=PQ$
Hnece proved

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#3 h Jun 2, 2014, 10:39 AM • 2 Y

Y by Adventure10, Mango247

We consider, $S=(O_1,r_1)$ and $T=(O_2,r_2)$ and we assume $r_1>r_2$ . Now draw, $O_2S\perp AP$ , clearly, $(O_2S)^2=4r_1r_2$ and $SP^2=(2r_1-r_2)^2$ ,
so, we have, $PQ^2=(PO_2)^2-(r_2)^2=(SO_2)^2+SP^2-(r_2)^2=4r_1r_2+(2r_1-r_2)^2-(r_2)^2=(4r_1)^2=AP^2$

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#4 h Jun 11, 2014, 10:32 AM • 2 Y

Y by Adventure10, Mango247

What do you denote by $S$ ?
First you assign them as circles and then you use $SP^2$

Am I missing something? (I sure am)

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#5 h Jun 11, 2014, 11:45 AM • 1 Y

Y by Adventure10

Omg! Just change the name of first circle instead of $S$

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#6 h Jun 11, 2014, 11:54 AM • 1 Y

Y by Adventure10

Then how do you define $S$ & $T$ ?

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#7 h Jun 11, 2014, 12:07 PM • 1 Y

Y by Adventure10

For example $SO_2$ here, $S$ is a point, if you change the name of the first circle $M$ instead of $S$ then the solution becomes clear!

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#8 h Jun 11, 2014, 12:21 PM • 2 Y

Y by Adventure10, Mango247

That's what I am asking!
Which point is $S$ ?

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#9 h Jun 11, 2014, 12:28 PM • 2 Y

Y by Adventure10, Mango247

Here $S$ is the foot of perp. To $AP$ from $O_2$ , actually I should select another name for this point because $S$ is already using as the name of the first circle, Got it?

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jlammy

#10 h Jul 4, 2014, 4:05 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Here is a solution using inversion.

Consider the circle $U$ with centre $P$ , and which passes through $A$ . Invert with respect to $U$ . The line $AB$ exchanges with circle $S$ . The line $PQ$ inverts to itself. The circle $T$ is tangent to the circle $S$ , the line $AB$ and the line $PQ$ , so $T$ inverts to itself. The point $Q$ inverts to a point on the line $PQ$ and also on the circle $T$ , so $Q$ inverts to $Q$ . Therefore, $PQ = AP$ , as required.

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