Given: OA = 10 cm, AB = 8 cm, AB ⊥ OP, PA is tangent to circle with centre O.
Step 1: Since PA is a tangent and OA is the radius, OA ⊥ PA (radius ⊥ tangent at point of contact).
Step 2: In right △OAB, AB ⊥ OB, so:
$$OB = \sqrt{OA^2 - AB^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 \text{ cm}$$
Step 3: In right △PAB (∠ABP = 90°):
$$PA = \sqrt{OA^2 - OB^2}$$
Wait — In right △OAP, OA ⊥ PA:
$$PA = \sqrt{OP^2 - OA^2}$$
First find OP: Since B lies on OP and OB = 6 cm, we need PB.
In right △PAB, PA ⊥ AB? No — PA is along the tangent. Since OA ⊥ PA and AB ⊥ OP:
In right △OAB: OB = 6 cm (shown above).
In right △PAB (∠ABP = 90°):
$$PB = \sqrt{PA^2 + AB^2}$$
In right △OAP: ∠OAP = 90°, so $OP^2 = OA^2 + PA^2$. Also $OP = OB + BP = 6 + PB$.
In right △PAB: $PA^2 = PB^2 - AB^2$ ... but ∠ABP = 90°, so $PB^2 = PA^2 + AB^2$.
From △OAP: $PA^2 = OP^2 - OA^2 = (6+PB)^2 - 100$
Substituting into $PB^2 = PA^2 + 64$:
$$PB^2 = (6 + PB)^2 - 100 + 64$$
$$PB^2 = PB^2 + 12\cdot PB + 36 - 36$$
$$0 = 12\cdot PB$$
This gives an error — rechecking: $PB^2 = (6+PB)^2 - 100 + 64$
$$PB^2 = PB^2 + 12PB + 36 - 36$$
$$0 = 12PB \Rightarrow$$ contradiction.
Correct approach — In △PAB, ∠PAB = 90° (since OA⊥PA and AB⊥OP means ∠PAB = 90°):
$$PB^2 = PA^2 + AB^2$$
From △OAP: $PA^2 = OP^2 - 100 = (6+PB)^2 - 100$
$$PB^2 = (6+PB)^2 - 100 + 64$$
$$PB^2 = PB^2 + 12PB + 36 - 36$$
$$0 = 12PB$$
Since ∠PAB = 90°, let me use △OAB correctly. OB = 6 cm; PB is along PO with B between P and O.
In right △PAB (∠PBA = 90°): $PA^2 = PB^2 + AB^2$... no, ∠PBA = 90° means $PA^2 = PB^2 + BA^2$.
$PA^2 = PB^2 + 64$ ... (i)
△OAP: $OP^2 = OA^2 + PA^2 \Rightarrow (OB+PB)^2 = 100 + PB^2 + 64$
$(6+PB)^2 = 164 + PB^2$
$36 + 12PB + PB^2 = 164 + PB^2$
$12PB = 128 \Rightarrow PB = \dfrac{128}{12} = \dfrac{32}{3}$ cm
$$\boxed{PB = \dfrac{32}{3} \approx 10.67 \text{ cm}}$$
Source: Chapter 10, Section 10.2 (Theorem 10.1)
---