Assertion (A): Two circles touch each other externally. The sum of their areas is 74π cm² and distance between their centres is 12 cm. Then difference between their radii is 2 cm.

Reason (R): When two circles touch each other internally, then two centres and point of contact lie on the same line and distance between their centres equals to the difference between two radii.

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

Both A and R are true.

Explanation

Let r₁ and r₂ be the radii of two circles.

Sum of their areas = 74π cm²

⇒ πr₁² + πr₂² = 74π

⇒ π(r₁² + r₂²) = 74π

⇒ r₁² + r₂² = 74 ……………..(1)

Distance between their centres = 12 cm.

⇒ r₁ + r₂ = 12

⇒ r₁ = 12 - r₂

Substituting r₁ = 12 - r₂ into equation (1),

⇒ (12 - r₂)² + r₂² = 74

⇒ (12)² + r₂² - 2 x 12 x r₂ + r₂² = 74

⇒ 144 + 2r₂² - 24r₂ = 74

⇒ 144 + 2r₂² - 24r₂ - 74 = 0

⇒ 2r₂² - 24r₂ + 70 = 0

⇒ r₂² - 12r₂ + 35 = 0

⇒ r₂² - (7r₂ + 5r₂) + 35 = 0

⇒ r₂² - 7r₂ - 5r₂ + 35 = 0

⇒ (r₂² - 7r₂) - (5r₂ - 35) = 0

⇒ r₂(r₂ - 7) - 5(r₂ - 7) = 0

⇒ (r₂ - 7)(r₂ - 5) = 0

⇒ r₂ = 7 or 5

If r₂ = 7, then r₁ = 12 - 7 = 5

If r₂ = 5, then r₁ = 12 - 5 = 7

The difference between the radii is 2 cm.

∴ Assertion (A) is true.

Given: Two circles with centers A and B touching each other internally at point P.

To Prove : P lies on the line AB produced.

Construction : Through the point of contact P, draw a common tangent PQ. Join AP and BP.

Proof : Angles between the radius and tangent are always 90°.

So, ∠ APQ = 90° and ∠ BPQ = 90°

∴ AP and BP both are perpendicular to the tangent PQ at the same point P.

Only one perpendicular can be drawn to a line through a point in it.

AP and BP lie in the same line.

⇒ ABP is a straight line.

∴ P lies on the line AB (when produced)

∴ Reason (R) is true.

Hence, both Assertion (A) and Reason (R) are true.