In figure 3.91, line PR touches the circle at point Q. Answer the following questions with the help of the figure.
(1)ATSQ is a cyclic quadrilateral.TAQ + TSQ = 180˚ [Opposite angles of cyclic quadrilateral are supplementary](2)Given PR is the tangent.Seg AQ is the secant.AQP = ½ m(arc AQ) [Angle between tangent and secant]ASQ = ½ m(arc AQ) [Inscribed angle]AQP ASQ(3)QTS = QAS [ Angles inscribed in same arc are equal]∠QTS = ½ m(arc QS) [Inscribed angle […]
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 […]
Draw circles with centres A, B and C each of radius 3 cm, such that each circle touches the other two circles.
Given radius of each circle = 3cm.If two circles touch each other externally, then the distance between their centres is equal to the sum of their radii.AB = 3+3 = 6AC = 3+3 = 6BC = 3+3 = 6Draw line segment AB = 6cm.With A as centre and radius = 6 cm, mark an arc.With […]
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 3.88, circles with centres X and Y touch internally at point Z . Seg BZ is a chord of bigger
Given X and Y are the centres of the circle.Proof:Join YZ In AXZ ,AX = XZ [Radii of same circle]XAZ = XZA….(i) [Isoceles triangle theorem]In BYZ,YB = YZ [Radii of same circle]YBZ = YZB [Isoceles triangle theorem]XZA = YBZ ..(ii) [Y-X-Z, B-A-Z]From (i) and (ii)XAZ = YBZIf a pair of corresponding angles formed by a […]
In the adjoining figure circles with centres X and Y touch each other at point Z. A secant passing
Construction: Draw segments XZ and YZProof: By theorem of touching circles, points X, Z, Y are collinear.XZA BZY vertically opposite anglesLet XZA = BZY = a ….. (I)Now, seg XA seg XZ …….. (Radii of same circle)XAZ = XZA. = a …….. (isosceles triangle theorem) (II)similarly, seg YB seg YZ (Radii of same circle)BZY = […]
In figure 3.86, circle with centre M touches the circle with centre N at point T.
(1) Given radius of bigger circle is 9cm.Length of segment MT = 9cm.(2) MT = MN+NT [M-N-T]9 = MN+2.5 [Given radius of smaller circle is 2.5]MN = 9-2.5MN = 6.5cm.(3) RM touches smaller circle at S. MR is the tangent to the smaller circle. NS is the radius of smaller circle.NSM = 90˚ [Tangent theorem]In […]
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 […]
In figure 3.84, O is the centre of the circle. Seg AB, seg AC are tangent segments. Radius of the circle
Construction: Draw OC and OB. Proof:Given AB and AC are the tangents and r is the radius of the circle.AB = AC …….(i) [Two tangents from a common point are congruent]OB = OC = r …(ii) [radii of same circle]Given AB = rAB = AC =OB = OC [From (i) and (ii)]OBA = OCA = […]