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denis_gylaev

Member

Registered: 2015-03-19

Posts: 66

More Circles

Point Y is on a circle and point P lies outside the circle such that \overline{PY} is tangent to the circle. Point A is on the circle such that segment \overline{PA} meets the circle again at point B. If PA = 15 and PY = 9, then what is AB?

Chords \overline{WY} and \overline{XZ} of a circle are perpendicular. If XV = 4, WV = 3, and VZ = 9, then find YZ.

Chords \overline{AB} and \overline{XY} of a circle meet at T. If XT = 4, TY = 6, and AT = 2TB, then what is AB?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: More Circles

hi denis_gylaev

There's a theorem in geometry that covers all three questions here.

I'll start with the middle question.  V is the point where WY and XZ cross.

In triangles XWV and YZV, angle XWY = XZY (using angle props. of a circle) and WVX = ZVY (= 90**) so the triangles are similar.  Thus:

In the third question the crossing point is outside the circle, but the theorem still holds.

In triangles TXA and TBY, angle T is common to both, angle XAB = XYB. So the triangles are similar and

Question 1.  If you rename point T as point P, then PB.PA = PY.PX

Let X approach Y.  Then the equation becomes, in the limit,  PB.PA = PY^2

Bob

**By vertically opposite angles this would still be true even if the angle wasn't 90.  But you need 90 in order to use Pythagoras theorem to complete the problem.

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