We at BYJUâ€™S provide class 10 Maths Chapter 10 Circles NCERT Solutions and are available in PDF format for free. Circles is one of the most important topic for the students of class 10 as well as for the various competitive board examinations. Students can either use this solution online or download the PDF files and prepare for their exams. These circles class 10 ncert solutions pdf are very detailed and prepared by subject our experts. You can see step-by-step solved solutions to all the questions existing in class 10 maths NCERT textbook syllabus (2018-2019).

This Circles chapter of class 10 maths deals with the various types of questions such as finding the angles, length of a chord, length of a tangent, tangent drawn from a point to a circle, etc., which should be practiced thoroughly to have a better learning experience and also to excel in the examination. BYJU’S study materials include different types of exemplar problems of class 10 on circles, radii of a circle, the diameter of a circle, finding the length of the segment and lot more.

### NCERT Solutions Class 10 Maths Chapter 10 Exercises

**1. In the figure, PQ is a chord of a circle and PT is the tangent at P such that âˆ QPT = 60Â°. Then find the measure of âˆ PRQ.**

**Solution.**

Since OP is **perpendicular** to PT.

âˆ OPT = 90Â°

âˆ OPQ = 90Â°- âˆ QPT

âˆ OPQ = 90 â€“ 60 = 30Â°.

In Î”OPQ, OP= OQ = r Â ( Radius of the circle )

âˆ OPQ= âˆ OQP = 30.

And,

âˆ POQ = 180 â€“ âˆ OPQ- âˆ OQP

= 180Â° â€“ 30Â° â€“ 30Â° = 120Â°

Also, reflex âˆ POQ = 360Â° â€“ 120Â° = 240Â°

Now, âˆ PRQ = Â reflex âˆ POQ

= 12x 240Â° = 120Â°

**2. If the angle between two radii of a circle is 130Â°, then find the degree measure of the angle between the tangents at the ends of the radii.**

**Solution.**

It is already known that angle between two radii and the angle between the tangents at the ends of the radii are supplementary.

Hence, Angle between the tangents at the ends of the radii is 180Â° â€“ 130Â° , i.e., 50Â°.

**3. In the figure, if âˆ AOB = 125Â°, then find the degree measure of âˆ **

**Solution.**

It is already known that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

âˆ AOB + âˆ COD = 180Â°.

125Â° + âˆ COD = 180Â°

âˆ COD = 180Â° â€“ 125Â° = 55Â°.

**4. In the figure, AB is a chord of the circle and AOC is its diameter such that âˆ ACB = 50Â°. If AT is the tangent to the circle at the point A, then find the measure of âˆ BAT.**

**Solution.**

âˆ BAT= 50Â°.

**5. At one end A of a diameter AB of a circle of radius 5cm, tangent XAY is drawn to the circle. Find the length of the chord CD parallel to XY and at a distance 8cm from A.**

**Solution.**

Since XAY is a tangent through one end A of a diameter of a circle,

AB is **perpendicular **to XY

And also, CD is **parallel **to XAY â‡’ AB is **perpendicular **to CD

Since OM is perpendicular from centre O to the chord CD.

â¤‘ OM is perpendicular bisector of chord CD.

That is., Â Â Â CM = MD = 12CD.

Now, AM =8cm (given)

â‡’ Â OM = AM â€“ AO = 8-5 = 3cm.

In OMC, we obtain

CM = CM=OC2âˆ’OM2âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆš

= 52âˆ’32âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆš

= 16âˆ’âˆ’âˆš

= 4 cm.

Hence, CD = 2cm = 2 x 4 = 8 cm.

**6. In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and âˆ OTA = 30Â°. Find the length of the segment AT.**

**Solution.**

In tight triangle âˆ†OTA, âˆ OTA = 30Â°.

OAOT=sin30âˆ˜

OA=12OT=12Ã—4=2cm

Again, AT=OT2âˆ’OA2âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆš

AT = 42âˆ’22âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆš

=16âˆ’4âˆ’âˆ’âˆ’âˆ’âˆ’âˆš

12âˆ’âˆ’âˆš

= 23â€“âˆšcm.

**7. In the figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50Â° with PQ, find the measure of âˆ POQ.**

**Solution.**

Since OP is **perpendicular **to PR

âˆ OPR = 90Â°

âˆ OPQ + âˆ QPR = 90Â°

âˆ OPQ + 50Â° = 90Â°

â†’ âˆ OPQ = âˆ OQP = 40Â°

Again,

âˆ POQ + âˆ OPQÂ + âˆ OQP = 180Â°

â†’ âˆ POQ + 40Â° + 40Â° = 180Â°

â†’ âˆ POQ = 180Â° â€“ 40Â° â€“ 40Â°

= 100Â°

**8. In the figure, if PA and PB are tangents to the circle with centre O such that âˆ APB = 50Â°, find the degree measure of âˆ OAB.**

**Solution.**

Since OA is **perpendicular** to PA and also, OB is **perpendicular **to PB

âˆ APB + âˆ AOB = 180Â°

50Â°+ âˆ AOB = 180Â°

âˆ AOB = 180Â° â€“Â 50Â° = 130Â°

In â–³AOB,

OA = OB = radii of same circle

âˆ OAB = âˆ OBA = x ( say )

Again, âˆ OAB + âˆ OBA + âˆ AOB = 180Â°

x +x + 130Â° = 180Â°

2x = 180Â° â€“Â 130Â° = 50Â°

X = 25Â°

Hence, âˆ OAB =25Â°

**9. In the figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and âˆ BQR = 70Â°, find the measure of âˆ AQB.**

**Solution.**

Since AB is **parallel** to PQR

âˆ B =âˆ BQR

â†’ âˆ B =70Â°

Also, âˆ A = âˆ BQRÂ Â Â [âˆ s in the corresponding alternate segments are equal ]

â†’ âˆ A = 70Â°

Now, in â–³ABQ, we have

âˆ A + âˆ B + âˆ AQB = 180Â°

70Â° + 70Â° + âˆ AQB = 180Â°

âˆ AQB=180Â° â€“ 70Â° â€“ 70Â°

=40Â°

Hence,Â âˆ AQB = 40Â°

**10. In the figure, DE and DF are tangents from an external point D to a circle with centre A. If DE=5cm and DE is perpendicular DF, find the radius of the circle.**

**Solution. **

Join AE and AF â†’ AE is **perpendicular **to DE and AF is **perpendicular **to DF.

AEDF is a square of side 5cm [DE = 5cm(given)]

Hence, radius of the circle is 5 cm.

**11. If two tangents inclined at an angle of 60Â° are drawn to a circle of radius 3cm, find the length of each tangent.**

**Solution.**

**.**

Here, let PA and PB be two tangents to a circle with centre O.

OA is **perpendicular** to AP.

Since OP bisects âˆ APB

âˆ OPA = 30Â°

Now, in â–³OAP, we have

APOA=cot30âˆ˜

â†’Â AP3=3â€“âˆš

â†’ AP=33â€“âˆšcm

**12. From a point P which is at a distance of 13cm from the centre O of a circle of radius 5cm, a pair of tangents PQ and PR to the circle are drawn. Find the area of the quadrilateral PQOR.**

**Solution.**

In â–³OPQ, we have

OQ **perpendicular **Â to PQ

OP2=PQ2+OQ2

132=PQ2+52

PQ2=169â€“25 = 144

PQ = 12cm

Now, (â–³OPQ) = x PQxOQ

= x12x5 = 30cm2

Thus, (quad, PQOR) = 2 x ar(â–³OPQ)

= 2Ã—30 cm2

= 60cm2