Chapter 10 · Circles

NCERT Class 10 Maths Exercise 10.2

All 13 questions of NCERT Class 10 Maths Exercise 10.2 (Circles), in the book's order, each with its answer and a step-by-step solution.

1. Q1
   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

   1. (a)7 cm
   2. (b)12 cm
   3. (c)15 cm
   4. (d)24.5 cm
   Show answer

   (a) 7 cm

   Step-by-step solution →
2. Q2
   In Fig. 10.11, ​if $TP$ and $TQ$ ​are the two ​tangents to a circle with centre $O$ so ​that $\angle POQ=110^{\circ }$, ​then $\angle PTQ$ is ​equal ​to

   1. (a)$60^{\circ }$
   2. (b)$70^{\circ }$
   3. (c)$80^{\circ }$
   4. (d)$90^{\circ }$
   Show answer

   (b) $70^{\circ }$

   Step-by-step solution →
3. Q3
   If tangents ​PA and PB from a ​point P to a ​circle with ​centre O ​are inclined to each other ​at angle of $80^{\circ }$, then $\angle POA$ is equal ​to

   1. (a)$50^{\circ }$
   2. (b)$60^{\circ }$
   3. (c)$70^{\circ }$
   4. (d)$80^{\circ }$
   Show answer

   (a) $50^{\circ }$

   Step-by-step solution →
4. Q4
   Prove ​that the tangents ​drawn at the ends ​of a ​diameter of ​a ​circle are parallel.
   Show answer

   Since $\angle PAB=\angle ABS=90^{\circ }$ and these are alternate angles made by the transversal $AB$, the tangents $PQ$ and $RS$ at the ends of the diameter $AB$ are parallel. Hence proved.

   Step-by-step solution →
5. Q5
   Prove ​that the perpendicular at the point ​of ​contact to ​the ​tangent to a ​circle ​passes through ​the centre.
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   The perpendicular to the tangent at the point of contact $P$ coincides with the radius $OP$, and therefore passes through the centre $O$. Hence proved.

   Step-by-step solution →
6. Q6
   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.
   Show answer

   Radius $r=3$ cm.

   Step-by-step solution →
7. Q7
   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.
   Show answer

   Length of the chord $=8$ cm.

   Step-by-step solution →
8. Q8
   A ​quadrilateral $ABCD$ ​is drawn to circumscribe a ​circle (see ​Fig. ​10.12). Prove ​that $AB+CD=AD+BC$.

   Show answer

   $AB+CD=AD+BC$. Hence proved.

   Step-by-step solution →
9. Q9
   In ​Fig. ​10.13, ​XY and ​X'Y' 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'Y' at B. Prove that $\angle AOB=90^{\circ }$.

   Show answer

   $\angle AOB=90^{\circ }$. Hence proved.

   Step-by-step solution →
10. Q10
    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 centre.
    Show answer

    $\angle APB+\angle AOB=180^{\circ }$, so the angle between the tangents is supplementary to the angle the chord of contact subtends at the centre. Hence proved.

    Step-by-step solution →
11. Q11
    Prove ​that ​the ​parallelogram circumscribing ​a circle is a ​rhombus.
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    All four sides are equal, $AB=BC=CD=DA$, so the circumscribing parallelogram $ABCD$ is a rhombus. Hence proved.

    Step-by-step solution →
12. Q12
    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.

    Show answer

    $AB=15$ cm and $AC=13$ cm.

    Step-by-step solution →
13. Q13
    Prove ​that ​opposite ​sides of ​a ​quadrilateral circumscribing a circle subtend ​supplementary angles at ​the centre of the circle.
    Show answer

    $\angle AOB+\angle COD=180^{\circ }$ and $\angle BOC+\angle AOD=180^{\circ }$, so opposite sides of the circumscribing quadrilateral subtend supplementary angles at the centre. Hence proved.

    Step-by-step solution →