Prove that any three points on a circle cannot be collinear.

Let O be centre of the circle. Let P, Q, R be any points on the circle.To prove: P,Q ,R cannot be collinear.Proof:OP = OQ [Radii of same circle] O is equidistant from end points P and Q of seg PQ.O lies on perpendicular bisector of PQ. [Perpendicular bisector theorem]In the same way we can […]

In figure 3.90, seg AB is a diameter of a circle with centre C. Line PQ is a tangent, which touches the circle at point T. seg AP line PQ and seg BQ line PQ. Prove that, seg CP seg CQ.

Given AB is the diameter of the circle with centre C.PQ is the tangent.AP PQBQPQTo prove seg CP seg CQConstruction: Join CP, CT and CQ. Proof:Since PQ is the tangent , CT PQAlso AP PQBQ PQAPCTBQ [Lines which are perpendicular to same lines are parallel]AC/CB = PT/TQ …(i) [Property of three parallel lines and their […]

In figure 3.89, line l touches the circle with centre O at point P. Q is the mid point of radius OP. RS is a chord through Q such that chords RS || line l. If RS = 12 find the radius of the circle.

Given line l is a tangent.Let the radius of circle be r.OP is the radius.OP line l. [Tangent theorem]Given chord RSline l.OP chord RSSince the perpendicular from centre of the circle to the chord bisects the chord,QS = ½ RSQS = ½ ×12 = 6OQ = r/2 [Given Q is the midpoint of OP]In OQSOS2 = OQ2+QS2 [Pythagoras theorem]r2 = (r/2)2+62r2-r2/4 = 36(3/4)r2 = […]

In figure 3.85, ABCD is a parallelogram. It circumscribes the circle with centre T.

Given ABCD is a parallelogram.AB = DC ……(i) [Opposite sides of parallelogram are equal]AD = BC ………(ii)AE = AH………(iii) [Two tangents from a common point are congruent]BE = BF……….(iv)CG = CF………..(v)DG = DH………..(vi)Adding (iii), (iv), (v),(vi)AE+BE+CG+DG = AH+BF+CF+DHAB+CD = AD+BC ……(vii) [AH+DH = AD, BF+CF = BC]From (i), (ii) and (vii)2AB = 2ADAB = ADAD […]