Question

PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

Hint:

Remember: \(AQ=AC\) and \(BR=BC\). This is the standard trick for problems involving triangles circumscribing circles or tangents intersecting.

Answer 1

Step 1: Using Power of a Point Theorem:
From external point P, two tangents PQ and PR are drawn.
Length of tangent = √(OP² − r²)

Given:
OP = 13 cm
Radius r = 5 cm

PQ = √(13² − 5²)
= √(169 − 25)
= √144
= 12 cm

Since tangents from same external point are equal:
PQ = PR = 12 cm

Step 2: Finding CP:
C lies on OP and OC is radius.
OC = 5 cm
OP = 13 cm

CP = OP − OC
= 13 − 5
= 8 cm

Step 3: Apply Power of Point from A:
From point A, AQ and AC are tangents.
Let AC = x cm.
Then AQ = x cm.

Since PQ = 12,
PA = PQ − AQ
= 12 − x

Now use right triangle ACP:
AP² = AC² + CP²

(12 − x)² = x² + 8²
(12 − x)² = x² + 64

Expand:
144 − 24x + x² = x² + 64

Cancel x² from both sides:
144 − 24x = 64

24x = 144 − 64
24x = 80
x = 80/24
x = 10/3 cm

Step 4: Finding Required Lengths:
AC = 10/3 cm

PA = 12 − 10/3
= (36 − 10)/3
= 26/3 cm

Since AC = BC (symmetry of tangents),
AB = AC + BC
= 10/3 + 10/3
= 20/3 cm

Final Answer:
AB = 20/3 cm
PA = 26/3 cm