{"id":119478,"date":"2022-05-04T16:15:31","date_gmt":"2022-05-04T10:45:31","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119478"},"modified":"2022-05-04T16:15:31","modified_gmt":"2022-05-04T10:45:31","slug":"chapter-10-circles-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-10-circles-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 10 – Circles Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

Exercise: 10.1<\/h2>\n

1. How many tangents can a circle have?<\/h2>\n

Answer:
\nThere can be\u00a0infinite\u00a0tangents to a circle. A circle is made up of infinite points which are at an equal distance from a point. Since there are infinite points on the circumference of a circle, infinite tangents can be drawn from them.<\/h3>\n

2. Fill in the blanks:
\n(i) A tangent to a circle intersects it in \u2026\u2026\u2026\u2026\u2026 point(s).
\n(ii) A line intersecting a circle in two points is called a \u2026\u2026\u2026\u2026.
\n(iii) A circle can have \u2026\u2026\u2026\u2026\u2026 parallel tangents at the most.
\n(iv) The common point of a tangent to a circle and the circle is called \u2026\u2026\u2026\u2026<\/h2>\n

Answer:
\n(i) A tangent to a circle intersects it in\u00a0one\u00a0point(s).
\n(ii) A line intersecting a circle in two points is called a\u00a0secant.
\n(iii) A circle can have\u00a0two\u00a0parallel tangents at the most.
\n(iv) The common point of a tangent to a circle and the circle is called the\u00a0point of contact.<\/h3>\n

3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at
\na point Q so that OQ = 12 cm. Length PQ is :
\n(A) 12 cm
\n(B) 13 cm
\n(C) 8.5 cm
\n(D) \u221a119 cm<\/h2>\n

Answer:<\/h3>\n

\"\"
\nIn the above figure, the line that is drawn from the centre of the given circle to the tangent PQ is perpendicular to PQ.
\nAnd so, OP \u22a5 PQ
\nUsing Pythagoras theorem in triangle \u0394OPQ we get,
\nOQ2\u00a0= OP2+PQ2
\n(12)2\u00a0= 52+PQ2
\nPQ2\u00a0= 144-25
\nPQ2\u00a0= 119
\nPQ = \u221a119 cm
\nSo,\u00a0option D\u00a0i.e. \u221a119 cm is the length of PQ.<\/h3>\n

4. Draw a circle and two lines parallel to a given line such that one is a tangent and the
\nother, a secant to the circle.<\/h2>\n

Answer:<\/h3>\n

\"\"
\nIn the above figure, XY and AB are two the parallel lines. The line segment AB is the tangent at point C while the line segment XY is the secant.<\/h3>\n

Exercise: 10.2 (Page NO: 213)<\/h2>\n

In Q.1 to 3, choose the correct option and give justification.
\n1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
\n(A) 7 cm
\n(B) 12 cm
\n(C) 15 cm
\n(D) 24.5 cm<\/h2>\n

Answer:
\nFirst, draw a perpendicular from the center O of the triangle to a point P on the circle which is touching the tangent. This line will be perpendicular to the tangent of the circle.<\/h3>\n

\"\"
\nSo, OP is perpendicular to PQ i.e. OP \u22a5 PQ
\nFrom the above figure, it is also seen that \u25b3OPQ is a right angled triangle.
\nIt is given that
\nOQ = 25 cm and PQ = 24 cm
\nBy using Pythagoras theorem in \u25b3OPQ,
\nOQ2\u00a0= OP2\u00a0+PQ2
\n(25)2\u00a0= OP2+(24)2
\nOP2\u00a0= 625-576
\nOP2\u00a0= 49
\nOP = 7 cm
\nSo, option A i.e. 7 cm is the radius of the given circle.<\/h3>\n

2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that \u2220POQ = 110\u00b0, then \u2220PTQ is equal to
\n(A) 60\u00b0
\n(B) 70\u00b0
\n(C) 80\u00b0
\n(D) 90\u00b0<\/h2>\n

Answer:
\nFrom the question, it is clear that OP is the radius of the circle to the tangent PT and OQ is the radius to the tangents TQ.<\/h3>\n

\"\"
\nSo, OP \u22a5 PT and TQ \u22a5 OQ
\n\u2234\u2220OPT = \u2220OQT = 90\u00b0
\nNow, in the quadrilateral POQT, we know that the sum of the interior angles is 360\u00b0
\nSo, \u2220PTQ+\u2220POQ+\u2220OPT+\u2220OQT = 360\u00b0
\nNow, by putting the respective values we get,
\n\u2220PTQ +90\u00b0+110\u00b0+90\u00b0 = 360\u00b0
\n\u2220PTQ = 70\u00b0
\nSo, \u2220PTQ is 70\u00b0 which is option B.<\/h3>\n

3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80\u00b0, then \u2220 POA is equal to
\n(A) 50\u00b0
\n(B) 60\u00b0
\n(C) 70\u00b0
\n(D) 80\u00b0<\/h2>\n

Answer:
\nFirst, draw the diagram according to the given statement.<\/h3>\n

\"\"
\nNow, in the above diagram, OA is the radius to tangent PA and OB is the radius to tangent PB.
\nSo, OA is perpendicular to PA and OB is perpendicular to PB i.e. OA \u22a5 PA and OB \u22a5 PB
\nSo, \u2220OBP = \u2220OAP = 90\u00b0
\nNow, in the quadrilateral AOBP,
\nThe sum of all the interior angles will be 360\u00b0
\nSo, \u2220AOB+\u2220OAP+\u2220OBP+\u2220APB = 360\u00b0
\nPutting their values, we get,
\n\u2220AOB + 260\u00b0 = 360\u00b0
\n\u2220AOB = 100\u00b0
\nNow, consider the triangles \u25b3OPB and \u25b3OPA. Here,
\nAP = BP (Since the tangents from a point are always equal)
\nOA = OB (Which are the radii of the circle)
\nOP = OP (It is the common side)
\nNow, we can say that triangles OPB and OPA are similar using SSS congruency.
\n\u2234\u25b3OPB \u2245 \u25b3OPA
\nSo, \u2220POB = \u2220POA
\n\u2220AOB = \u2220POA+\u2220POB
\n2 (\u2220POA) = \u2220AOB
\nBy putting the respective values, we get,
\n=>\u2220POA = 100\u00b0\/2 = 50\u00b0
\nAs angle \u2220POA is 50\u00b0 option A is the correct option.<\/h3>\n

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.<\/h2>\n

Answer:
\nFirst, draw a circle and connect two points A and B such that AB becomes the diameter of the circle. Now, draw two tangents PQ and RS at points A and B respectively.<\/h3>\n

\"\"
\nNow, both radii i.e. AO and OB are perpendicular to the tangents.
\nSo, OB is perpendicular to RS and OA perpendicular to PQ
\nSo, \u2220OAP = \u2220OAQ = \u2220OBR = \u2220OBS = 90\u00b0
\nFrom the above figure, angles OBR and OAQ are alternate interior angles.
\nAlso, \u2220OBR = \u2220OAQ and \u2220OBS = \u2220OAP (Since they are also alternate interior angles)
\nSo, it can be said that line PQ and the line RS will be parallel to each other. (Hence Proved).<\/h3>\n

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.<\/h2>\n

Solution:
\nLet, O is the centre of the given circle.
\nA tangent PR has been drawn touching the circle at point P.
\nDraw QP \u22a5 RP at point P, such that point Q lies on the circle.<\/h3>\n

\"\"
\n\u2220OPR = 90\u00b0 (radius \u22a5 tangent)
\nAlso, \u2220QPR = 90\u00b0 (Given)
\n\u2234 \u2220OPR = \u2220QPR
\nNow, the above case is possible only when centre O lies on the line QP.
\nHence, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.<\/h3>\n

6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.<\/h2>\n

Answer:
\nDraw the diagram as shown below.<\/h3>\n

\"\"
\nHere, AB is the tangent that is drawn on the circle from a point A.
\nSo, the radius OB will be perpendicular to AB i.e. OB \u22a5 AB
\nWe know, OA = 5cm and AB = 4 cm
\nNow, In \u25b3ABO,
\nOA2\u00a0=AB2+BO2\u00a0(Using Pythagoras theorem)
\n52\u00a0= 42+BO2
\nBO2\u00a0= 25-16
\nBO2\u00a0= 9
\nBO = 3
\nSo, the radius of the given circle i.e. BO is 3 cm.<\/h3>\n

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.<\/h2>\n

Answer:
\nDraw two concentric circles with the center O. Now, draw a chord AB in the larger circle which touches the smaller circle at a point P as shown in the figure below.<\/h3>\n

\"\"
\nFrom the above diagram, AB is tangent to the smaller circle to point P.
\n\u2234 OP \u22a5 AB
\nUsing Pythagoras theorem in triangle OPA,
\nOA2= AP2+OP2
\n52\u00a0= AP2+32
\nAP2\u00a0= 25-9
\nAP = 4
\nNow, as OP \u22a5 AB,
\nSince the perpendicular from the center of the circle bisects the chord, AP will be equal to PB
\nSo, AB = 2AP = 2\u00d74 = 8 cm
\nSo, the length of the chord of the larger circle is 8 cm.<\/h3>\n

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC<\/h2>\n

Answer:
\nThe figure given is:<\/h3>\n

\"\"
\nFrom this figure we can conclude a few points which are:
\n(i) DR = DS
\n(ii) BP = BQ
\n(iii) AP = AS
\n(iv) CR = CQ
\nSince they are tangents on the circle from points D, B, A, and C respectively.
\nNow, adding the LHS and RHS of the above equations we get,
\nDR+BP+AP+CR = DS+BQ+AS+CQ
\nBy rearranging them we get,
\n(DR+CR) + (BP+AP) = (CQ+BQ) + (DS+AS)
\nBy simplifying,
\nAD+BC= CD+AB<\/h3>\n

9. In Fig. 10.13, XY and X\u2032Y\u2032 are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X\u2032Y\u2032 at B. Prove that \u2220 AOB = 90\u00b0.<\/h2>\n

Answer:
\nFrom the figure given in the textbook, join OC. Now, the diagram will be as-<\/h3>\n

\"\"
\nNow the triangles \u25b3OPA and \u25b3OCA are similar using SSS congruency as:
\n(i) OP = OC They are the radii of the same circle
\n(ii) AO = AO It is the common side
\n(iii) AP = AC These are the tangents from point A
\nSo, \u25b3OPA \u2245 \u25b3OCA
\nSimilarly,
\n\u25b3OQB \u2245 \u25b3OCB
\nSo,
\n\u2220POA = \u2220COA \u2026 (Equation i)
\nAnd, \u2220QOB = \u2220COB \u2026 (Equation ii)
\nSince the line POQ is a straight line, it can be considered as a diameter of the circle.
\nSo, \u2220POA +\u2220COA +\u2220COB +\u2220QOB = 180\u00b0
\nNow, from equations (i) and equation (ii) we get,
\n2\u2220COA+2\u2220COB = 180\u00b0
\n\u2220COA+\u2220COB = 90\u00b0
\n\u2234\u2220AOB = 90\u00b0<\/h3>\n

10. Prove 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 center.<\/h2>\n

Answer:
\nFirst, draw a circle with centre O. Choose an external point P and draw two tangents PA and PB at point A and point B respectively. Now, join A and B to make AB in a way that it subtends \u2220AOB at the center of the circle. The diagram is as follows:<\/h3>\n

\"\"
\nFrom the above diagram, it is seen that the line segments OA and PA are perpendicular.
\nSo, \u2220OAP = 90\u00b0
\nIn a similar way, the line segments OB \u22a5 PB and so, \u2220OBP = 90\u00b0
\nNow, in the quadrilateral OAPB,
\n\u2234\u2220APB+\u2220OAP +\u2220PBO +\u2220BOA = 360\u00b0 (since the sum of all interior angles will be 360\u00b0)
\nBy putting the values we get,
\n\u2220APB + 180\u00b0 + \u2220BOA = 360\u00b0
\nSo, \u2220APB + \u2220BOA = 180\u00b0 (Hence proved).<\/h3>\n

11. Prove that the parallelogram circumscribing a circle is a rhombus.<\/h2>\n

Answer:
\nConsider a parallelogram ABCD which is circumscribing a circle with a center O. Now, since ABCD is a parallelogram, AB = CD and BC = AD.<\/h3>\n

\"\"
\nFrom the above figure, it is seen that,
\n(i) DR = DS
\n(ii) BP = BQ
\n(iii) CR = CQ
\n(iv) AP = AS
\nThese are the tangents to the circle at D, B, C, and A respectively.
\nAdding all these we get,
\nDR+BP+CR+AP = DS+BQ+CQ+AS
\nBy rearranging them we get,
\n(BP+AP)+(DR+CR) = (CQ+BQ)+(DS+AS)
\nAgain by rearranging them we get,
\nAB+CD = BC+AD
\nNow, since AB = CD and BC = AD, the above equation becomes
\n2AB = 2BC
\n\u2234 AB = BC
\nSince AB = BC = CD = DA, it can be said that ABCD is a rhombus.<\/h3>\n

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.<\/h2>\n

Answer:
\nThe figure given is as follows:<\/h3>\n

\"\"
\nConsider the triangle ABC,
\nWe know that the length of any two tangents which are drawn from the same point to the circle is equal.
\nSo,
\n(i) CF = CD = 6 cm
\n(ii) BE = BD = 8 cm
\n(iii) AE = AF =\u00a0x
\nNow, it can be observed that,
\n(i) AB = EB+AE = 8+x
\n(ii) CA = CF+FA = 6+x
\n(iii) BC = DC+BD = 6+8 = 14
\nNow the semi perimeter \u201cs\u201d will be calculated as follows
\n2s = AB+CA+BC
\nBy putting the respective values we get,
\n2s = 28+2x
\ns = 14+x<\/h3>\n

\"\"
\nBy solving this we get,
\n= \u221a(14+x)48x\u00a0\u2026\u2026\u2026 (i)
\nAgain, the area of \u25b3ABC = 2 \u00d7 area of (\u25b3AOF + \u25b3COD + \u25b3DOB)
\n= 2\u00d7[(\u00bd\u00d7OF\u00d7AF)+(\u00bd\u00d7CD\u00d7OD)+(\u00bd\u00d7DB\u00d7OD)]
\n= 2\u00d7\u00bd(4x+24+32) = 56+4x\u00a0\u2026\u2026\u2026\u2026..(ii)
\nNow from (i) and (ii) we get,
\n\u221a(14+x)48x\u00a0= 56+4x
\nNow, square both the sides,
\n48x(14+x) = (56+4x)2
\n48x =\u00a0[4(14+x)]2\/(14+x)
\n48x =\u00a016(14+x)
\n48x =\u00a0224+16x
\n32x =\u00a0224
\nx =\u00a07 cm
\nSo, AB = 8+x
\ni.e. AB = 15 cm
\nAnd, CA = x+6 =13 cm.<\/h3>\n

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.<\/h2>\n

Answer:
\nFirst draw a quadrilateral ABCD which will circumscribe a circle with its centre O in a way that it touches the circle at point P, Q, R, and S. Now, after joining the vertices of ABCD we get the following figure:<\/h3>\n

\"\"
\nNow, consider the triangles OAP and OAS,
\nAP = AS (They are the tangents from the same point A)
\nOA = OA (It is the common side)
\nOP = OS (They are the radii of the circle)
\nSo, by SSS congruency \u25b3OAP \u2245 \u25b3OAS
\nSo, \u2220POA = \u2220AOS
\nWhich implies that\u22201 = \u22208
\nSimilarly, other angles will be,
\n\u22204 = \u22205
\n\u22202 = \u22203
\n\u22206 = \u22207
\nNow by adding these angles we get,
\n\u22201+\u22202+\u22203 +\u22204 +\u22205+\u22206+\u22207+\u22208 = 360\u00b0
\nNow by rearranging,
\n(\u22201+\u22208)+(\u22202+\u22203)+(\u22204+\u22205)+(\u22206+\u22207) = 360\u00b0
\n2\u22201+2\u22202+2\u22205+2\u22206 = 360\u00b0
\nTaking 2 as common and solving we get,
\n(\u22201+\u22202)+(\u22205+\u22206) = 180\u00b0
\nThus, \u2220AOB+\u2220COD = 180\u00b0
\nSimilarly, it can be proved that \u2220BOC+\u2220DOA = 180\u00b0
\nTherefore, the opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the center of the circle.<\/h3>\n","protected":false},"excerpt":{"rendered":"

Class 10 Mathematics NCERT book solutions for Chapter 10 – Circles Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":117421,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-119478","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119478","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=119478"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119478\/revisions"}],"predecessor-version":[{"id":119502,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119478\/revisions\/119502"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/117421"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=119478"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=119478"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=119478"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}