PQ is a chord of a circle. The tangent XR at point X on the circle intersects the extension of PQ at point R. Given that XR = 12 cm, PQ = x cm, and QR = (x - 2) cm, find the value of x.

Question

Question

This question was previously asked in

SSC CHSL Exam 2024 Tier-I Official Paper (Held On: 05 Jul, 2024 Shift 2)

Answer 1

Given:

XR = 12 cm (the length of the tangent)

PQ = x cm (the length of the chord)

QR = x - 2 cm (the external part of the secant)

Formula used:

According to the Power of a Point Theorem:

\(XR^2 = PR \times QR\)

Where PR = PQ + QR = (x + (x - 2)) = 2x - 2

Substitute into the formula:

Calculations:

12² = (2x - 2)(x - 2)

⇒ 144 = 2x(x - 2) - 2(x - 2)

⇒ 144 = 2x² - 4x - 2x + 4

⇒ 144 = 2x² - 6x + 4

Move everything to one side:

⇒ 2x² - 6x + 4 - 144 = 0

⇒ 2x² - 6x - 140 = 0

Simplify the equation by dividing through by 2:

⇒ x² - 3x - 70 = 0

Factoring the quadratic equation:

(x - 10)(x + 7) = 0

So, x = 10 or x = -7

Conclusion:

Since x represents the length of PQ, it must be positive, so we discard x = -7.

∴ x = 10 cm.