A point P is outside a circle with radius 8 units. A secant line passing through P and the circle’s center O intersects the circle at points A and B, where B is the point on the secant closer to P. A tangent from P touches the circle at T, and PT=15 units. Find the length of the segment PB.
9 units
Given: radius \(r = 8\) units, tangent \(PT = 15\) units. B is nearer to P and A is farther, and A, B, O are collinear (the secant passes through the center).
By the tangent–secant relation (power of a point): \(PT^{2} = PB \times PA\).
Since the secant passes through the centre O, AB is a diameter, so \(AB = 2r = 16\) units.
Let \(PB = x\). Then \(PA = PB + AB = x + 16\).
Substitute: \(15^{2} = x(x+16)\), i.e. \(225 = x^{2} + 16x\).
So \(x^{2} + 16x - 225 = 0\).
Solving: \(x = \dfrac{-16 + \sqrt{16^{2} + 4\times225}}{2} = \dfrac{-16 + \sqrt{1156}}{2} = \dfrac{-16 + 34}{2} = 9\).
Hence the length of segment PB is 9 units.
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