Page 1 Circle 1) A point A is 25 cm from the centre of the circle. The length of the tangent drawn fromA to the circle is 24 cm. Find the radius. Solution: AB is a tangent to the circle with point of contact B and OB is the radius. ? OB AB …… (Tangent at any point of the circle is perpendicular to the radius.) OBA= 90 ( Radius is perpendicular to the tangent.) In ?OAB, 90 OAB ? ? ? (Pythagoras Theorem) AO 2 = AB 2 + OB 2 2 2 2 25 24 OB ? ? ? 2 625 576 OB ?? 2 625 576 OB?? A 2 49 OB ? Taking square root of both the sides we get, 7 OB ? cm The radius of the circle is 7cm 2) In the fig, if AB=AC, prove that BE=EC Solution: AB is tangent to the circle with the point of contact D, AC is tangent to the circle with the point of contact F and BC is tangent to the circle with the point of contact E. AB=AC----------------------(given) Tangents drawn from an exterior point to the circle are equal. ?AD=AF........... (1) BD=BE............ (2) CF=CE............ (3) AD+BD=AF+FC------- (since AB = AC and D & F are midpoints of AB & AC respectively.) AF+BD=AF+FC (from (1)) ?BD=FC.......... (4) From (2), (3) and (4) BE=CE 3) If tangents AB and AC from a point A to a circle are inclined to each other at 80 ? then find BOC ? Solution: Page 2 Circle 1) A point A is 25 cm from the centre of the circle. The length of the tangent drawn fromA to the circle is 24 cm. Find the radius. Solution: AB is a tangent to the circle with point of contact B and OB is the radius. ? OB AB …… (Tangent at any point of the circle is perpendicular to the radius.) OBA= 90 ( Radius is perpendicular to the tangent.) In ?OAB, 90 OAB ? ? ? (Pythagoras Theorem) AO 2 = AB 2 + OB 2 2 2 2 25 24 OB ? ? ? 2 625 576 OB ?? 2 625 576 OB?? A 2 49 OB ? Taking square root of both the sides we get, 7 OB ? cm The radius of the circle is 7cm 2) In the fig, if AB=AC, prove that BE=EC Solution: AB is tangent to the circle with the point of contact D, AC is tangent to the circle with the point of contact F and BC is tangent to the circle with the point of contact E. AB=AC----------------------(given) Tangents drawn from an exterior point to the circle are equal. ?AD=AF........... (1) BD=BE............ (2) CF=CE............ (3) AD+BD=AF+FC------- (since AB = AC and D & F are midpoints of AB & AC respectively.) AF+BD=AF+FC (from (1)) ?BD=FC.......... (4) From (2), (3) and (4) BE=CE 3) If tangents AB and AC from a point A to a circle are inclined to each other at 80 ? then find BOC ? Solution: Circle AB and AC are tangents to the circle with the point of contact B and C respectively. OB and OC are radii of the circle.(OB AB, OC AC) 90 OCA ? ? ? ?& 90 OBA ? ? ? ? 80 BAC ? ? ? ------------------------ (Given) In Quadrilateral OBAC 360 OBA BAC OCA BOC ? ? ? ? ? ? ? ? ? (Sum of angles in a quadrilateral) 90 80 90 360 BOC ? ? ? ? ? ? ? ? ? ? ? 360 90 80 90 BOC ? ? ? ? ? ? ? ? ? 360 260 100 BOC ? ? ? ? ? ? ? 100 BOC ? ? ? ? 4) Prove that in two concentric circles, the chord of larger circle, which touches the smaller circle, is bisected at the point of contact of the smaller circle. Solution: Let O be the centre of the two concentric circles and let PQ be the chord of the bigger circle which touch the smaller circle at the point M. Join OM. For the smaller circle, OM is radius and PQ is tangent. OM PQ ?? (radius is perpendicular to the tangent at the point of contact) For the bigger circle, OM is line drawn from the centre perpendicular to the chord PQ PM QM ?? M bisects the chord PQ ? 5) In the fig, AB=3cm, is the tangent to the circle (O, 2cm), then find the length of BC. Solution: AB is the tangent to the circle(O,2cm) Since r=2cm ? d=4cm i.e. AC = 4cm.----(diameter=2 radius) We know that OA is the radius and AB is the tangent. 90 OAB ? ? ? ?(tangent at any point of the circle is perpendicular to the radius OA AB) Page 3 Circle 1) A point A is 25 cm from the centre of the circle. The length of the tangent drawn fromA to the circle is 24 cm. Find the radius. Solution: AB is a tangent to the circle with point of contact B and OB is the radius. ? OB AB …… (Tangent at any point of the circle is perpendicular to the radius.) OBA= 90 ( Radius is perpendicular to the tangent.) In ?OAB, 90 OAB ? ? ? (Pythagoras Theorem) AO 2 = AB 2 + OB 2 2 2 2 25 24 OB ? ? ? 2 625 576 OB ?? 2 625 576 OB?? A 2 49 OB ? Taking square root of both the sides we get, 7 OB ? cm The radius of the circle is 7cm 2) In the fig, if AB=AC, prove that BE=EC Solution: AB is tangent to the circle with the point of contact D, AC is tangent to the circle with the point of contact F and BC is tangent to the circle with the point of contact E. AB=AC----------------------(given) Tangents drawn from an exterior point to the circle are equal. ?AD=AF........... (1) BD=BE............ (2) CF=CE............ (3) AD+BD=AF+FC------- (since AB = AC and D & F are midpoints of AB & AC respectively.) AF+BD=AF+FC (from (1)) ?BD=FC.......... (4) From (2), (3) and (4) BE=CE 3) If tangents AB and AC from a point A to a circle are inclined to each other at 80 ? then find BOC ? Solution: Circle AB and AC are tangents to the circle with the point of contact B and C respectively. OB and OC are radii of the circle.(OB AB, OC AC) 90 OCA ? ? ? ?& 90 OBA ? ? ? ? 80 BAC ? ? ? ------------------------ (Given) In Quadrilateral OBAC 360 OBA BAC OCA BOC ? ? ? ? ? ? ? ? ? (Sum of angles in a quadrilateral) 90 80 90 360 BOC ? ? ? ? ? ? ? ? ? ? ? 360 90 80 90 BOC ? ? ? ? ? ? ? ? ? 360 260 100 BOC ? ? ? ? ? ? ? 100 BOC ? ? ? ? 4) Prove that in two concentric circles, the chord of larger circle, which touches the smaller circle, is bisected at the point of contact of the smaller circle. Solution: Let O be the centre of the two concentric circles and let PQ be the chord of the bigger circle which touch the smaller circle at the point M. Join OM. For the smaller circle, OM is radius and PQ is tangent. OM PQ ?? (radius is perpendicular to the tangent at the point of contact) For the bigger circle, OM is line drawn from the centre perpendicular to the chord PQ PM QM ?? M bisects the chord PQ ? 5) In the fig, AB=3cm, is the tangent to the circle (O, 2cm), then find the length of BC. Solution: AB is the tangent to the circle(O,2cm) Since r=2cm ? d=4cm i.e. AC = 4cm.----(diameter=2 radius) We know that OA is the radius and AB is the tangent. 90 OAB ? ? ? ?(tangent at any point of the circle is perpendicular to the radius OA AB) Circle 90 CAB ? ? ? ? (C-A-B)---------(C – O – A) In CAB, CAB= 90 2 2 2 BC AB AC ? ? ? (Pythagoras theorem) 2 2 2 34 BC ?? 2 9 16 25 BC ? ? ? Taking square roots on both sides we get, BC=5cm 6) Prove that the tangents drawn at the end points of a diameter of a circle are parallel. Solution: Consider a circle with centre at ‘O’ and radius ‘R’ [i.e. (O, R)]. Let MN be the diameter of the circle(O, R) and let AB and PQ are the tangents through the end points of the diameter MN i.e. through the points ‘M’ & ‘N’ respectively OM and ON are radii of the circle and AB and PQ are tangents at M and N respectively. OM AB & ON PQ 90 OMB ? ? ? ? and 90 ONQ ? ? ? ? 90 90 180 ONQ OMB ? ? ? ? ? ? ? ? ? ? These angles are remote interior angles. Since the remote Interior angles are supplementary the lines are parallel. Hence, AB is parallel to PQ. 7) In two concentric circles, the radii are 5cm and 13 cm. If the chord of the circle (O, 13cm) is the tangent of circle (O, 5cm) then find the length of the chord of the bigger circle. Solution:To find AB = ? Radius of the smaller circle =OC=r=5cm Radius of the bigger circle=OA=R=13cm For the smaller circle, OC=OP=5cm AB is tangent to smaller circle and OP is radius. OP AB ?? And AP=PB------------------ (1) Consider ?OAP, in this OPA=90 2 2 2 OA OP AP ?? (Pythagoras Theorem) Page 4 Circle 1) A point A is 25 cm from the centre of the circle. The length of the tangent drawn fromA to the circle is 24 cm. Find the radius. Solution: AB is a tangent to the circle with point of contact B and OB is the radius. ? OB AB …… (Tangent at any point of the circle is perpendicular to the radius.) OBA= 90 ( Radius is perpendicular to the tangent.) In ?OAB, 90 OAB ? ? ? (Pythagoras Theorem) AO 2 = AB 2 + OB 2 2 2 2 25 24 OB ? ? ? 2 625 576 OB ?? 2 625 576 OB?? A 2 49 OB ? Taking square root of both the sides we get, 7 OB ? cm The radius of the circle is 7cm 2) In the fig, if AB=AC, prove that BE=EC Solution: AB is tangent to the circle with the point of contact D, AC is tangent to the circle with the point of contact F and BC is tangent to the circle with the point of contact E. AB=AC----------------------(given) Tangents drawn from an exterior point to the circle are equal. ?AD=AF........... (1) BD=BE............ (2) CF=CE............ (3) AD+BD=AF+FC------- (since AB = AC and D & F are midpoints of AB & AC respectively.) AF+BD=AF+FC (from (1)) ?BD=FC.......... (4) From (2), (3) and (4) BE=CE 3) If tangents AB and AC from a point A to a circle are inclined to each other at 80 ? then find BOC ? Solution: Circle AB and AC are tangents to the circle with the point of contact B and C respectively. OB and OC are radii of the circle.(OB AB, OC AC) 90 OCA ? ? ? ?& 90 OBA ? ? ? ? 80 BAC ? ? ? ------------------------ (Given) In Quadrilateral OBAC 360 OBA BAC OCA BOC ? ? ? ? ? ? ? ? ? (Sum of angles in a quadrilateral) 90 80 90 360 BOC ? ? ? ? ? ? ? ? ? ? ? 360 90 80 90 BOC ? ? ? ? ? ? ? ? ? 360 260 100 BOC ? ? ? ? ? ? ? 100 BOC ? ? ? ? 4) Prove that in two concentric circles, the chord of larger circle, which touches the smaller circle, is bisected at the point of contact of the smaller circle. Solution: Let O be the centre of the two concentric circles and let PQ be the chord of the bigger circle which touch the smaller circle at the point M. Join OM. For the smaller circle, OM is radius and PQ is tangent. OM PQ ?? (radius is perpendicular to the tangent at the point of contact) For the bigger circle, OM is line drawn from the centre perpendicular to the chord PQ PM QM ?? M bisects the chord PQ ? 5) In the fig, AB=3cm, is the tangent to the circle (O, 2cm), then find the length of BC. Solution: AB is the tangent to the circle(O,2cm) Since r=2cm ? d=4cm i.e. AC = 4cm.----(diameter=2 radius) We know that OA is the radius and AB is the tangent. 90 OAB ? ? ? ?(tangent at any point of the circle is perpendicular to the radius OA AB) Circle 90 CAB ? ? ? ? (C-A-B)---------(C – O – A) In CAB, CAB= 90 2 2 2 BC AB AC ? ? ? (Pythagoras theorem) 2 2 2 34 BC ?? 2 9 16 25 BC ? ? ? Taking square roots on both sides we get, BC=5cm 6) Prove that the tangents drawn at the end points of a diameter of a circle are parallel. Solution: Consider a circle with centre at ‘O’ and radius ‘R’ [i.e. (O, R)]. Let MN be the diameter of the circle(O, R) and let AB and PQ are the tangents through the end points of the diameter MN i.e. through the points ‘M’ & ‘N’ respectively OM and ON are radii of the circle and AB and PQ are tangents at M and N respectively. OM AB & ON PQ 90 OMB ? ? ? ? and 90 ONQ ? ? ? ? 90 90 180 ONQ OMB ? ? ? ? ? ? ? ? ? ? These angles are remote interior angles. Since the remote Interior angles are supplementary the lines are parallel. Hence, AB is parallel to PQ. 7) In two concentric circles, the radii are 5cm and 13 cm. If the chord of the circle (O, 13cm) is the tangent of circle (O, 5cm) then find the length of the chord of the bigger circle. Solution:To find AB = ? Radius of the smaller circle =OC=r=5cm Radius of the bigger circle=OA=R=13cm For the smaller circle, OC=OP=5cm AB is tangent to smaller circle and OP is radius. OP AB ?? And AP=PB------------------ (1) Consider ?OAP, in this OPA=90 2 2 2 OA OP AP ?? (Pythagoras Theorem) Circle 2 2 2 13 5 AP ?? 2 169 25 AP ?? 2 169 25 AP?? 2 144 AP ? Taking square root on both sides we get, AP=12cm AB=2AP=2 x 12--- [ AB = AP + PB & by (1)] 24 AB cm ?? 8) In the fig, AB and CD are two parallel tangents of a circle whose centre is O. Another tangent PQ at X intersects AB at Y and CD at Z. Prove that 90 YOZ ? ? ? Solution: Tangents AB and CD are parallel. Tangent PQ intersects AB and CD at Y and Z respectively.(Let 1 P ? ? ? 2 AOY ? ? ? 3 YOX ? ? ? ) To Prove: 90 YOZ ? ? ? 4 XOZ ? ? ? 5 ZOC ? ? ? 6 C ? ? ? Construction: Join OA,OC and OX. Proof: AB and PQ are tangents to the circle and OA and OX are radii to the circle. OA AB and OX PQ. OAB = OXP 1 6 90 ? ? ? ? ? ? (Radius is perpendicular to the tangent)........(1) Consider right ? YAO and right ?YXO, we have OAB = OXP 1 6 90 ? ? ? ? ? (from (1)) OY=OY (Common side) OA=OX (Radii of the same circle) ??YAO ? ?YXO (Hypotenuse-side Theorem= two right triangles are congruent if hypotenuses and corresponding side is congruent.) AOY = YOX 23 ? ? ? ? (C.A.C.T)......(2) Similarly we can prove ??ZCO ? ?ZXO XOZ = COZ 45 ? ? ? ? (C.A.C.T)........(3) But 2 3 4 5 180 ? ? ? ? ? ? ? ? ? (Linear Pair) From (2) and (3) we get, 3 3 4 4 180 ? ? ? ? ? ? ? ? ? ? ? 2 3 4 180 ? ? ? ? ? 3 4 90 ? ? ? ? ? 90 YOZ ? ? ? ? Page 5 Circle 1) A point A is 25 cm from the centre of the circle. The length of the tangent drawn fromA to the circle is 24 cm. Find the radius. Solution: AB is a tangent to the circle with point of contact B and OB is the radius. ? OB AB …… (Tangent at any point of the circle is perpendicular to the radius.) OBA= 90 ( Radius is perpendicular to the tangent.) In ?OAB, 90 OAB ? ? ? (Pythagoras Theorem) AO 2 = AB 2 + OB 2 2 2 2 25 24 OB ? ? ? 2 625 576 OB ?? 2 625 576 OB?? A 2 49 OB ? Taking square root of both the sides we get, 7 OB ? cm The radius of the circle is 7cm 2) In the fig, if AB=AC, prove that BE=EC Solution: AB is tangent to the circle with the point of contact D, AC is tangent to the circle with the point of contact F and BC is tangent to the circle with the point of contact E. AB=AC----------------------(given) Tangents drawn from an exterior point to the circle are equal. ?AD=AF........... (1) BD=BE............ (2) CF=CE............ (3) AD+BD=AF+FC------- (since AB = AC and D & F are midpoints of AB & AC respectively.) AF+BD=AF+FC (from (1)) ?BD=FC.......... (4) From (2), (3) and (4) BE=CE 3) If tangents AB and AC from a point A to a circle are inclined to each other at 80 ? then find BOC ? Solution: Circle AB and AC are tangents to the circle with the point of contact B and C respectively. OB and OC are radii of the circle.(OB AB, OC AC) 90 OCA ? ? ? ?& 90 OBA ? ? ? ? 80 BAC ? ? ? ------------------------ (Given) In Quadrilateral OBAC 360 OBA BAC OCA BOC ? ? ? ? ? ? ? ? ? (Sum of angles in a quadrilateral) 90 80 90 360 BOC ? ? ? ? ? ? ? ? ? ? ? 360 90 80 90 BOC ? ? ? ? ? ? ? ? ? 360 260 100 BOC ? ? ? ? ? ? ? 100 BOC ? ? ? ? 4) Prove that in two concentric circles, the chord of larger circle, which touches the smaller circle, is bisected at the point of contact of the smaller circle. Solution: Let O be the centre of the two concentric circles and let PQ be the chord of the bigger circle which touch the smaller circle at the point M. Join OM. For the smaller circle, OM is radius and PQ is tangent. OM PQ ?? (radius is perpendicular to the tangent at the point of contact) For the bigger circle, OM is line drawn from the centre perpendicular to the chord PQ PM QM ?? M bisects the chord PQ ? 5) In the fig, AB=3cm, is the tangent to the circle (O, 2cm), then find the length of BC. Solution: AB is the tangent to the circle(O,2cm) Since r=2cm ? d=4cm i.e. AC = 4cm.----(diameter=2 radius) We know that OA is the radius and AB is the tangent. 90 OAB ? ? ? ?(tangent at any point of the circle is perpendicular to the radius OA AB) Circle 90 CAB ? ? ? ? (C-A-B)---------(C – O – A) In CAB, CAB= 90 2 2 2 BC AB AC ? ? ? (Pythagoras theorem) 2 2 2 34 BC ?? 2 9 16 25 BC ? ? ? Taking square roots on both sides we get, BC=5cm 6) Prove that the tangents drawn at the end points of a diameter of a circle are parallel. Solution: Consider a circle with centre at ‘O’ and radius ‘R’ [i.e. (O, R)]. Let MN be the diameter of the circle(O, R) and let AB and PQ are the tangents through the end points of the diameter MN i.e. through the points ‘M’ & ‘N’ respectively OM and ON are radii of the circle and AB and PQ are tangents at M and N respectively. OM AB & ON PQ 90 OMB ? ? ? ? and 90 ONQ ? ? ? ? 90 90 180 ONQ OMB ? ? ? ? ? ? ? ? ? ? These angles are remote interior angles. Since the remote Interior angles are supplementary the lines are parallel. Hence, AB is parallel to PQ. 7) In two concentric circles, the radii are 5cm and 13 cm. If the chord of the circle (O, 13cm) is the tangent of circle (O, 5cm) then find the length of the chord of the bigger circle. Solution:To find AB = ? Radius of the smaller circle =OC=r=5cm Radius of the bigger circle=OA=R=13cm For the smaller circle, OC=OP=5cm AB is tangent to smaller circle and OP is radius. OP AB ?? And AP=PB------------------ (1) Consider ?OAP, in this OPA=90 2 2 2 OA OP AP ?? (Pythagoras Theorem) Circle 2 2 2 13 5 AP ?? 2 169 25 AP ?? 2 169 25 AP?? 2 144 AP ? Taking square root on both sides we get, AP=12cm AB=2AP=2 x 12--- [ AB = AP + PB & by (1)] 24 AB cm ?? 8) In the fig, AB and CD are two parallel tangents of a circle whose centre is O. Another tangent PQ at X intersects AB at Y and CD at Z. Prove that 90 YOZ ? ? ? Solution: Tangents AB and CD are parallel. Tangent PQ intersects AB and CD at Y and Z respectively.(Let 1 P ? ? ? 2 AOY ? ? ? 3 YOX ? ? ? ) To Prove: 90 YOZ ? ? ? 4 XOZ ? ? ? 5 ZOC ? ? ? 6 C ? ? ? Construction: Join OA,OC and OX. Proof: AB and PQ are tangents to the circle and OA and OX are radii to the circle. OA AB and OX PQ. OAB = OXP 1 6 90 ? ? ? ? ? ? (Radius is perpendicular to the tangent)........(1) Consider right ? YAO and right ?YXO, we have OAB = OXP 1 6 90 ? ? ? ? ? (from (1)) OY=OY (Common side) OA=OX (Radii of the same circle) ??YAO ? ?YXO (Hypotenuse-side Theorem= two right triangles are congruent if hypotenuses and corresponding side is congruent.) AOY = YOX 23 ? ? ? ? (C.A.C.T)......(2) Similarly we can prove ??ZCO ? ?ZXO XOZ = COZ 45 ? ? ? ? (C.A.C.T)........(3) But 2 3 4 5 180 ? ? ? ? ? ? ? ? ? (Linear Pair) From (2) and (3) we get, 3 3 4 4 180 ? ? ? ? ? ? ? ? ? ? ? 2 3 4 180 ? ? ? ? ? 3 4 90 ? ? ? ? ? 90 YOZ ? ? ? ? Circle 9) In the fig, two circles touch each other at C. A line is drawn through C to meet the circles at A and B respectively. Prove that tangents drawn at A and B are parallel. Solution: Since the two circles touch each other at the point C. The line through the two centres will pass through the point of contact. O C Q ? ? ? // O-C-O’ Construction: join OA and O’B. Consider ?OAC, OA=OC (Radii of the same circle) OAC OCA ? ? ? ? (Isosceles triangle theorem)........(1) Now Consider ?O’CB O’C=O’B (Radii of the same circle) '' O CB O BC ? ? ? ? (Isosceles triangle theorem)..........(2) But, ' O CB OCA ? ? ? (Vertically Opposite Angles)...........(3) From (1), (2) and (3) we get, ' O BC OAC ? ? ? ' O BA OAB ? ? ? ? (A-C-B).....(4) Consider Circle with centre O AP is the tangent and OA is the radius. OA AP 90 OAP ? ? ? ? Similarly ' 90 O BQ ? ? ? ? ' O BQ OAP ? ? ? ? (Both 90 ? ) ' O BA QBA OAB PAB ? ? ? ? ? ? ? ? (Angle Addition Property) '' O BA QBA O BA PAB ? ? ? ? ? ? ? (From (4)) Cancelling ' O BA ? on both sides we get, QBA PAB ? ? ? By alternate angle test we get, AP BQ AP BQ ? 10) In the fig, A is a point of intersection of two circles with centres O and O’. The tangent at A meets the circles at B and C respectively again. Now let P be a point so that AOPO’ is a parallelogram. Prove that P is the circumcentre of ?ABC. Solution: Let R and r be the radii of the circles with the centres O & O’ respectively. AC is tangent to the circle (O,R) AB is tangent to the circle (O’, r)Read More

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