Properties of Circles Part 4

Tangent from Exterior point


In the figure on the above, P is a point outside the circle, with centre O, PA and PB are two tangents drawn from P to touch the circle at A and B respectively. We can find that

i) AP = BP
ii) ÐAPO = ÐBPO
iii) ÐAOP = ÐBOP

ÐOAP = ÐOBP = 90° (tan ⊥ rad.)
△AOP and △BOP are congruent (RHS Property)
AP = BP
ÐAPO = ÐBPO and ÐAOP = ÐBOP

We can conclude that:

a) tangents drawn to a circle from an external point are equal

b) the tangents subtend equal angles at the centre

c) the line joining the external point to the centre of the circle bisects the angle between the tangents.

The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Thus, ÐPTB = ÐPQT.

Alternate Segment Theorem


The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Thus, ÐPTB = ÐPQT.

Proof:

ÐSPT = 90° (rt. Ð in a semicircle)
Ðx + Ðy = 90° ( Ð sum of ∆)
Ðy + Ðz = 90° ( tan rad.)
Ðx = Ðz
ÐPTB = ÐPST
ÐPST = ÐPQT ( Ðs in same segment)
ÐPTB = ÐPQT

Example 1:

In the figure shown, PA and PB are tangents to the circle. ABP = 60° and BAC = 40°. Find the value of x and y.

x° = 60° (Ð in alt. segment)
ÐPAB = 40° (base Ð of isos. ∆PAB)
y° = 180° - 60° - 40° (adj. Ðs on str. l)
= 80°

x is 60 and y is 80.