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Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is(a) 10 units (b) 20 units (c) 12 units (d) 24 units |
Answer» (d) 24 units AB ⊥ chord PQ ⇒ AB bisects chord PQ ⇒ PQ = 2PB. AB = \(\sqrt{(2-5)^2+(-3-1)^2}\) = \(\sqrt{(-3)^2+(-4)^2}\) = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5 AP = radius of circle = 13 ∴ By Pythagoras’ Theorem, PB = \(\sqrt{AP^2-AB^2}\) = \(\sqrt{169-25}\) = \(\sqrt{144}\) = 12 units ∴ PQ = 2 × PB = 24 units. |
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