Question

PT is tangent to Oo. Secant BA is perpendicular to PT at P.If TA-6 and PA-3, find (a) AB (b) the distance from 0 to AB (c) the radius of O 0.

Answer 1

. Given: PT tangent to circle o at pont Secant BA is perpendicular to PT at P. TA=6 and PA-3 L Consider DAPT right angled at P CAP geld =) (PT)2 = (AT)? (AP) = (6)2-(3)2 = 36 -9 = 27 & PT = 127 - (PT = 383 -0. Now, (PT-PT = PA-PB (Tangent - secant theorem) (PT)2 = (PA)(PA + B) (By segment addihem podulole) (313)2 = 3 (3 + AB) (from ) 27 = 3+ AB & AB = 9-3 AB= 6) Au A LOTP = 900 (Radice draws to point of Tangency in S I to the tangent LOTP = [TPM = {PMO = 90°

Also LOTP + LTPM + LAMO+ (MOT = 360$ (sumo interior angles Ja quodsilokial) » Mot = 360-07P - LTPM - LPmo = 360 - (90*4 90°90°) • 360-270 → Mor - 90° tlenur, in quadrlateref OTPM, au do inferior angles are equal to 90, therefore OTPm is in general a rectangle (sith opposite sides eguop and parallel) So, om = PT com = 3.83] (distance from 0 to AB) c) Mis mich point of chord AB became line passing through the center and perpendicular to a chanel bisects the chood. = AM = MB DAB) = (6) => AM - MB = 3] - @ Now, consider some right angled at M: (OB)? com) t (MB)2 (Pythagoras Theorem) = (3/3)2+3)2 = 27+9 (from b) and egn®) 7 OB = √36 = | OB = 6) (radius of oo.)