9 uestion. Exercise 9. How many tangents can a circle have? Solution For every point of a circle, we can draw a tangent. Therefore, infinite tangents can be drawn. uestion. Fill in the blanks. (i) tangent to a circle intersects it in... point(s). (ii) line intersecting a circle in two points is called a.... (iii) circle can have... parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called.... Solution (i) ne. tangent line touch or intersect the circle only at one point. (ii) Secant. ny line intersecting the circle at two points is called a secant. (iii) Two. Maximum a circle an have two parallel tangents which can be drawn to the opposite side of the centre. (iv) oint of contact. uestion 3. tangent at a point of a circle of radius 5 cm meets a line through the centre at a point so that = cm. Length is (a) cm (b) 3 cm (c) 8.5 cm (d) 9 cm Solution (d) Here, is a tangent, which touches a point of the circle. nd = 5 cm is a radius of circle. Join = cm. Since, radius of circle is perpendicular to the tangent. In right, y ythagoras theorem = + ( ) = () 5 + = 44 5 = 9 = 9 cm uestion 4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle. Solution Firstly draw a circle with centre and draw a line l. Now, we draw a two parallel lines to l, such that one line m is tangent to the circle and another n is secant to the circle. 5 cm cm m n l
9 Exercise 9. Choose the correct option and give justification. uestion. From a point, the length of the tangent to a circle is 4 cm and the distance of from the centre is 5 cm. The radius of the circle is (a) 7 cm (b) cm (c) 5 cm (d) 4.5 cm Solution (a) Given that, the length of the tangent = 4 cm and = 5 cm. Join. 4 cm 5 cm We know, radius is perpendicular to the tangent. In right = + (y ythagoras theorem) 5 = + 4 = 65 576 = 49 = 7 cm uestion. In figure, if T and T are the two tangents to a circle with centre so that = 0, then T is equal to T 0 (a) 60 (b) 70 (c) 80 (d) 90 Solution (b) Join, T 0
In = = radius of circle = (Equal sides have corresponding equal angles) 80 0 = = = 35 Since, T and T (Radius of circle is perpendicular to the tangent) T = 90 and T = 90 + T = 90 and + T = 90 35 + T = 90 and 35 + T = 90 T = 55 and T = 55 In T, T + T + T = 80 55 + 55 + T = 80 T = 80 ( 55 + 55 )= 70 uestion 3. If tangents and from a point to a circle with centre are inclined to each other at angle of 80, then is equal to (a) 50 (b) 60 (c) 70 (d) 80 Solution (a) and are two tangents drawn from a point. 80 ie.., = 80 Join. Since, tangents drawn from external point to the circle is equal in lengths. ie.., = = (i) In, + + = 80 80 + = 80 [From Eq. (i)] = 00 = 50 = = 50 Since, and = 90 and = 90 + = 90 and + = 90 + =
50 + = 90 and 50 + = 90 = 40 and = 40 Now, in, + + = 80 40 + 40 + = 80 = 80 80 = 00 Since, angle bisector divides an angle into two equal parts. = = 00 = 50 uestion 4. rove that the tangents drawn at the ends of a diameter of a circle are parallel. Solution Let be a diameter of a given circle and let LM and be the tangent lines drawn to the circle at points and, respectively. Since, the tangent at a point to a circle is perpendicular to the radius through the point. L M and LM = 90 and M = 90 = M LM Hence proved. uestion 5. rove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. Solution Tangent MN is drawn from point of the circle. perpendicular line is drawn to the MN line. R 90 M ie.., M = 90 Take any point R on the circle. N
Join R and R. y using alternate segment theorem M = R = 90 We know, any chord subtends an angle 90 only when chord is a diameter of the circle. Hence, perpendicular is always passes through the centre. uestion 6. The length of a tangent from a point at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. Solution Given, = 5 cm and M = 4 cm 4 cm M 5 cm M M (Radius is perpendicular to M) In right M, = M + M (y ythagoras theorem) 5 = M + 4 M = 5 6= 9 M = 3 cm Hence, radius of the circle is 3 cm. uestion 7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Solution Here, we draw two circles C and C of radii r = 3 cm and r = 5 cm Now, we draw a chord such that it touches the circle C at point D. The centre of concentric circle is. Now, we draw a perpendicular bisector from to which meets at D. C C 3 cm D 5 cm ie.., D = D In right D, = D + D (y ythagoras theorem)
5 = 3 + D D = 5 9 = 6 D = 4 cm Length of chord = = D = 4 = 8 cm uestion 8. quadrilateral CD is drawn to circumscribe a circle (see figure). rove that + CD = D + C. D S Solution Using theorem, the lengths of tangents drawn from an external point to a circle are equal. Suppose, is an external point, then = S (i) Suppose, is an external point, then = (ii) Suppose, C is an external point, then C = RC (iii) Suppose, D is an external point, then SD = RD (iv) n adding Eqs. (i), (ii), (iii) and (iv), we get ( + ) + ( RC + RD) = ( S + ) + ( C + SD) + CD = ( S + SD) + ( + C) + CD = D + C Hence proved. uestion 9. In figure, XY and X Y are two parallel tangents to a circle with centre and another tangent with point of contact C intersecting XY at and XY at. rove that = 90. X R C Y C X' Y'
Solution Since, tangents drawn from an external point to a circle are equal. = C Thus, in and C, = C = (Common) = C (Radius of circle) y SSS criterion of congruence, we have C = C C = C Similarly, we can prove that C = C = C Since, XY X Y C + C = 80 (Sum of interior angles on the same side of transversal is 80 ) C + C = 80 (i) C + C = 90 In, C + C + = 80 C + C = 80 (ii) From Eqs. (i) and (ii), we get 80 = 90 = 90 uestion 0. rove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. Solution Let and R be two tangents drawn from an external point to a circle with centre. We have to prove that, R = 80 R or R + R = 80 In right and R, R = R (Tangents drawn from an external point are equal) = R (Radius of circle) = (Common)
Therefore, by SSS criterion of congruence, R = R and = R R = and R = (i) In, + = 90 = 90 = 80 (Multiplying both sides by ) R = 80 R [From Eq. (i)] R + R = 80 Hence proved. uestion. rove that the parallelogram circumscribing a circle is a rhombus. Solution Let CD be a parallelogram circumscribing a circle. Using the property that the tangents to a circle from an exterior point are equal in length. D C N M = and M = N C = CN and D = D n adding all equations, we get ( M + M) + ( C + D) = + N + CN + D + CD = ( + D) + ( N + NC)= D + C = C (CD is a parallelogram. = C = C = CD = D Hence, CD is a rhombus. Hence proved. M = M + M, C = N + NC, CD = C + D, D = + D = CD andc = D)
uestion. C is drawn to circumscribe a circle of radius 4 cm such that the segments D and DC into which C is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides and C. C D 6 cm 8 cm Solution Given, CD = 6 cm, D = 8 cm and radius = 4 cm x x 6 F 4 4 E 8 C D 6 cm 8 cm Join C, and. y using the property, tangents drawn from external point equal in length. CD = CF = 6 cm and D = E = 8 cm Let F = E = x cm In C, rea of triangle, = ase Height = C D = 4 4 = 8cm In C, rea of triangle, = C F = + = + ( 6 x) 4 x
In, rea of triangle, 3 = E x = ( 8 + ) 4 = 6 + x Now, perimeter of triangle, C = + C + C ( ) S = ( x + 6 + 4 + 8 + x) S = 4 + Now, area of C = + + 3 = 8 + ( + x) + ( 6 + x) Using Heron's formula, rea of C = s ( s a)( s b)( s c) x = 56 + 4x (i) = ( 4 + x)( 4 + x 4)( 4 + x x 6)( 4 + x x 8) = ( 4 + x)( x)( 8)() 6 = ( 4 + x) x 48 From Eqs. (i) and (ii), we get n squaring both sides, ( 4 + x) x 48 = 56 + 4x ( 4 + x) 48 x = 4 ( 4 + x) 3x = 4 + x x = 4 x = 7 Length C = 6 + x = 6 + 7 = 3 cm Length of = 8 + x = 8 + 7 = 5 cm (ii) uestion 3. rove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Solution Let CD be a quadrilateral circumscribing a circle with centre. circle touches the sides of a quadrilateral at points E, F, G and H. D F C G 7 6 8 5 4 3 E H
To prove + CD = 80 and D + C = 80 Construction Join H, E, F and G. roof Using the property, two tangents drawn from an external point to a circle subtend equal angles at the centre. = 3 = 4 5 = 6 7 = 8 (i) We know the sum of all angles subtended at a point is 360. + + 3 + 4 + 5 + 6+ 7 + 8 = 360 ( + 3 + 6+ 7) = 360 and ( + 8 + 4 + 5) = 360 [From Eq. (i)] ( + 3) + ( 6 + 7) = 80 and ( + 8) + ( 4 + 5) = 80 + CD = 80 and D + C = 80 Hence proved.