# Thread: Circles and tangents....

1. ## Circles and tangents....

The problem:
Two circles touch one another externally at point A. A straight line through A meets one of the circles at P and the other at Q. Prove that the tangent to the first circle at P is parallel to the tangent to the second circle at Q.

I don't even know how to begin this ...any help would be appreciated!!

2. I don't know how to draw here, so draw the figure on paper.

Draw two circles (no need to be same in size) that kiss each other at point A.
Call the center of one circle, B, and the center of the other circle, C. Draw line segment BC.
Draw line segment PQ such that it passes through point A, it ends on the other side of circle B at point P, and it ends also at the other side of circle C at point Q.
Draw radii BP and CQ.
Draw the tangent lines at points P and Q.
Extend radius PB to meet the (tangent line at Q) at point R.
Extend radius QC to meet the (tangent line at P) at point S.

Now we are set.

Angles BAP and CAQ are equal, because they are vertical angles. ----(1)

Triangle PBA is isosceles, because radius BP = radius BA.
So, angle BPA = angle BAP ------------------------------------(2)

Triangle QCA is isosceles, because radius CQ = radius CA.
So, angle CQA= angle CAQ ------------------------------------(3)

Play with (1),(2),(3),
Hence, angle BPA = angle CQA ------------------(4)

In circle B:
radius BP is perpendicular to the tangent line PS, so angle BPS is 90 degrees.
And so, angle APS = 90deg minus angle BPA --------------(5)

In circle C:
radius CQ is perpendicular to the tangent line QR, so angle CQR is 90 degrees.
And so, angle AQR = 90deg minus angle CQA --------------(6)

Play with (4),(5),(6),
Hence, angle APS = angle AQR

Angles APS and AQR are alternate interior angles of tangents PS and QR, with line segment PQ as the transversal line.

Therefore, since angles APS and AQR are equal, then tangents PS and QR are parallel to each other. ----------proven.

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If you don't like that reasoning, we can show that angles PRQ and QSP are 90 degrees each. So PRQS is a rectangle. So RQ is parallel to PS.

3. Hello, star_109!

A variation of ticbol's excellent proof . . .
. . (and I'll use his diagram.)

Two circles touch one another externally at point A.
A straight line through A meets one of the circles at P and the other at Q.
Prove that the tangent to the first circle at P
is parallel to the tangent to the second circle at Q.
Let $T_P$ denote the tangent at $P$, and $T_Q$ the tangent at $Q$.

$\Delta PBA$ and $\Delta QCA$ are isosceles. .Each has equal radii for two of its sides.
. . Hence: . $\angle BPA = \angle BAP$ and $\angle CAQ = \angle CQA$

But $\angle BAP = \angle CAQ$ . . . . vertical angles
. . Hence: . $\angle BPA = \angle CQA$

Then: . $BP \parallel CQ$ . . . . alternate-interior angles

A radius is perpendicular to a tangent at the point of tangency.
. . Hence: $T_P \perp BP$ and $T_Q \perp CQ$

Therefore: . $T_P \parallel T_Q$