Chapter 9 · Circles

NCERT Exemplar Class 10 Maths Exercise 9.3

All 10 questions of NCERT Exemplar Class 10 Maths Exercise 9.3 (Circles), in the book's order, each with its answer and a step-by-step solution.

1. Q1
   Out ​of the two ​concentric ​circles, ​the radius of ​the outer circle is 5 cm and ​the chord $AC$ of length 8 cm ​is a tangent to the inner circle. Find the radius ​of the inner circle.
   Show answer

   The radius of the inner circle is $OM=\sqrt{25-16}=3$ cm.

   Step-by-step solution →
2. Q2
   Two ​tangents PQ and PR ​are drawn from an ​external point to a circle with centre ​O. Prove that ​QORP is ​a ​cyclic quadrilateral.
   Show answer

   $\angle OQP+\angle ORP=90^{\circ }+90^{\circ }=180^{\circ }$, so QORP is a cyclic quadrilateral.

   Step-by-step solution →
3. Q3
   If from an ​external ​point B of a circle with centre O, ​two ​tangents BC and BD ​are drawn such that $\angle DBC=120^{\circ }$, prove that $BC+BD=BO$, i.e., $BO=2BC$.
   Show answer

   $BC+BD=2BC=BO$, proved.

   Step-by-step solution →
4. Q4
   Prove ​that the centre of a ​circle touching ​two intersecting ​lines lies on the angle ​bisector ​of the lines.
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   Since $△OAP\cong △OBP$ (RHS), $\angle OPA=\angle OPB$, so $OP$ bisects the angle and the centre $O$ lies on the angle bisector.

   Step-by-step solution →
5. Q5
   In Fig. ​9.13, ​AB and CD ​are common tangents ​to two circles of unequal radii. ​Prove ​that $AB=CD$.

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   $AB=CD$ (proved using equal tangents from the external point $E$).

   Step-by-step solution →
6. Q6
   In Question ​5 above, if radii ​of the ​two circles are ​equal, prove ​that AB $=$ CD.
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   Let $O_1$ and $O_2$ be the centres and $r$ the common radius. A radius is perpendicular to the tangent at the point of contact, so $O_{1}A\perp AB$ and $O_{2}B\perp AB$, with $O_{1}A=O_{2}B=r$. Hence $O_{1}AB{O}_2$ is a rectangle and $AB=O_1{O}_2$. In the same way $O_{1}CD{O}_2$ is a rectangle and $CD=O_1{O}_2$. Therefore $AB=CD$.

   Step-by-step solution →
7. Q7
   In Fig. 9.14, ​common ​tangents $AB$ and $CD$ to ​two circles ​intersect ​at $E$. Prove that $AB=CD$.

   Show answer

   $AB=CD$ (proved using equal tangents from $E$).

   Step-by-step solution →
8. Q8
   A ​chord ​PQ of a circle is parallel to ​the tangent drawn at a point R of the circle. Prove ​that ​R bisects ​the ​arc PRQ.
   Show answer

   R bisects the arc PRQ: the tangent-chord angle theorem with the parallel condition gives $\angle RPQ=\angle RQP$, so $RP=RQ$ and therefore arc $RP=$ arc $RQ$.

   Step-by-step solution →
9. Q9
   Prove that the ​tangents drawn at ​the ​ends of ​a ​chord ​of a circle make equal angles with the chord.
   Show answer

   Since $PA=PB$ (equal tangents from $P$), triangle $PAB$ is isosceles, so $\angle PAB=\angle PBA$ — the tangents make equal angles with the chord.

   Step-by-step solution →
10. Q10
    Prove that ​a diameter AB of a circle bisects all those ​chords which are ​parallel ​to ​the ​tangent ​at ​the point A.
    Show answer

    Proved: the tangent at A is perpendicular to diameter AB; any chord CD parallel to it is therefore perpendicular to AB, and since AB passes through the centre, it bisects CD.

    Step-by-step solution →