NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2#

Let O be the centre and QT the tangent, with T the point of contact. Then QT = 24 cm and OQ = 25 cm.

OT is the radius and QT is the tangent at T, so OT ⊥ QT.

In right △OTQ, by Pythagoras Theorem:

$$OT^2 = OQ^2 - QT^2$$

$$\Rightarrow OT^2 = 25^2 - 24^2$$

$$\Rightarrow OT^2 = 625 - 576$$

$$\Rightarrow OT^2 = 49$$

$$\Rightarrow OT = 7$$

∴ Answer: (A) 7 cm

OP and OQ are radii, and TP and TQ are tangents at P and Q, so:
$\angle OPT = 90°$
$\angle OQT = 90°$

In quadrilateral OPTQ, the angles add up to 360°:

$$\angle POQ + \angle OPT + \angle PTQ + \angle OQT = 360°$$

$$\Rightarrow 110° + 90° + \angle PTQ + 90° = 360°$$

$$\Rightarrow \angle PTQ = 70°$$

∴ Answer: (B) 70°

In △OAP and △OBP:
OA = OB (radii)
PA = PB (tangents from the external point P)
OP = OP (common)

So by SSS congruence:

$$\triangle OAP \cong \triangle OBP$$

$$\Rightarrow \angle APO = \angle BPO$$

$\angle APB = 80°$, so:

$$\angle APO = \frac{80°}{2} = 40°$$

OA is the radius and PA is the tangent at A, so $\angle OAP = 90°$.

In △OAP, the angles add up to 180°:

$$\angle POA + \angle OAP + \angle APO = 180°$$

$$\Rightarrow \angle POA + 90° + 40° = 180°$$

$$\Rightarrow \angle POA = 50°$$

∴ Answer: (A) 50°

Given: AB is a diameter of a circle with centre O. PQ is the tangent at A and RS is the tangent at B, with P and R on the same side of AB.

To prove: PQ ∥ RS

Proof:

OA is the radius and PQ is the tangent at A, so:

$$\angle PAB = 90° \quad \cdots (1)$$

OB is the radius and RS is the tangent at B, so:

$$\angle ABS = 90° \quad \cdots (2)$$

From equation (1) and equation (2):

$$\angle PAB = \angle ABS$$

These are alternate angles formed by the transversal AB with the lines PQ and RS.

∴ PQ ∥ RS. Hence proved. $\blacksquare$

Given: A circle with centre O, and the tangent PQ touching it at A.

To prove: The perpendicular to PQ at A passes through O.

Proof:

OA is the radius and PQ is the tangent at A, so OA ⊥ PQ.

Suppose the perpendicular to PQ at A does not pass through O. Then it is a different line AX with AX ⊥ PQ.

So both OA and AX are perpendicular to PQ at the same point A. But only one perpendicular can be drawn to a line at a given point on it, so this is impossible.

∴ The perpendicular to PQ at A is the line AO, so it passes through the centre O. Hence proved. $\blacksquare$

Let O be the centre, AT the tangent with point of contact T, and $r$ the radius. Then OA = 5 cm and AT = 4 cm.

OT is the radius and AT is the tangent at T, so OT ⊥ AT.

In right △OTA, by Pythagoras Theorem:

$$OT^2 = OA^2 - AT^2$$

$$\Rightarrow r^2 = 5^2 - 4^2$$

$$\Rightarrow r^2 = 9$$

$$\Rightarrow r = 3$$

∴ The radius of the circle is 3 cm.

Let O be the common centre, and let AB be the chord of the larger circle that touches the smaller circle at M.

AB is a tangent to the smaller circle at M, so OM ⊥ AB, with OM = 3 cm and OA = 5 cm.

In right △OMA, by Pythagoras Theorem:

$$AM^2 = OA^2 - OM^2$$

$$\Rightarrow AM^2 = 25 - 9$$

$$\Rightarrow AM^2 = 16$$

$$\Rightarrow AM = 4$$

The perpendicular from the centre to a chord bisects the chord, so:

$$AB = 2 \times AM = 8$$

∴ The length of the chord is 8 cm.

Let the circle touch AB, BC, CD and DA at P, Q, R and S respectively.

Tangents drawn from an external point are equal, so:

$$AP = AS \quad \cdots (1)$$

$$BP = BQ \quad \cdots (2)$$

$$CR = CQ \quad \cdots (3)$$

$$DR = DS \quad \cdots (4)$$

Adding equation (1), equation (2), equation (3) and equation (4):

$$\begin{array}{rcl} AP &=& AS \\ BP &=& BQ \\ CR &=& CQ \\ DR &=& DS \\ \hline AP + BP + CR + DR &=& AS + BQ + CQ + DS \end{array}$$

$$\Rightarrow (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$$

$$\Rightarrow AB + CD = AD + BC$$

Hence proved. $\blacksquare$

Let XY touch the circle at P and X′Y′ touch it at Q. Join OP, OQ and OC.

In △OPA and △OCA:
OP = OC (radii)
AP = AC (tangents from the external point A)
OA = OA (common)

So by SSS congruence, △OPA ≅ △OCA, and:

$$\angle PAO = \angle CAO$$

$$\Rightarrow \angle CAO = \frac{1}{2}\angle PAB \quad \cdots (1)$$

In the same way, △OQB ≅ △OCB, so:

$$\angle QBO = \angle CBO$$

$$\Rightarrow \angle CBO = \frac{1}{2}\angle QBA \quad \cdots (2)$$

XY ∥ X′Y′ and AB is a transversal, so the co-interior angles add up to 180°:

$$\angle PAB + \angle QBA = 180° \quad \cdots (3)$$

Adding equation (1) and equation (2), and using equation (3):

$$\angle CAO + \angle CBO = \frac{1}{2}(\angle PAB + \angle QBA)$$

$$\Rightarrow \angle CAO + \angle CBO = 90°$$

In △AOB, the angles add up to 180°:

$$\angle AOB + \angle CAO + \angle CBO = 180°$$

$$\Rightarrow \angle AOB + 90° = 180°$$

$$\Rightarrow \angle AOB = 90°$$

Hence proved. $\blacksquare$

Let PA and PB be the tangents from an external point P to a circle with centre O, with A and B the points of contact.

OA and OB are radii, and PA and PB are tangents at A and B, so:
$\angle OAP = 90°$
$\angle OBP = 90°$

In quadrilateral OAPB, the angles add up to 360°:

$$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360°$$

$$\Rightarrow \angle AOB + 90° + \angle APB + 90° = 360°$$

$$\Rightarrow \angle AOB + \angle APB = 180°$$

∴ The angle between the tangents, ∠APB, is supplementary to the angle at the centre, ∠AOB. Hence proved. $\blacksquare$

Let ABCD be a parallelogram circumscribing a circle.

From Question 8, for a quadrilateral circumscribing a circle:

$$AB + CD = AD + BC \quad \cdots (1)$$

ABCD is a parallelogram, so its opposite sides are equal:

$$CD = AB \quad \cdots (2)$$

$$BC = AD \quad \cdots (3)$$

Substituting equation (2) and equation (3) in equation (1):

$$AB + CD = AD + BC$$

$$AB + AB = AD + AD$$

$$\Rightarrow 2AB = 2AD$$

$$\Rightarrow AB = AD$$

So AB = BC = CD = DA.

∴ All four sides of the parallelogram are equal, so ABCD is a rhombus. Hence proved. $\blacksquare$

Let the circle, with centre O, touch AB at F and AC at E. Let AF = $x$ cm.

Tangents drawn from an external point are equal, so:
BF = BD = 8 cm
CE = CD = 6 cm
AE = AF = $x$ cm

So the sides are:
AB = $x + 8$
AC = $x + 6$
BC = 8 + 6 = 14

Area using Heron’s formula: The semi-perimeter is:

$$s = \frac{(x + 8) + (x + 6) + 14}{2} = x + 14$$

So we find ar(△ABC):

$$= \sqrt{s(s - BC)(s - AC)(s - AB)}$$

$$= \sqrt{(x + 14)(x)(8)(6)}$$

$$= \sqrt{48x(x + 14)}$$

Area using the radius: OD, OE and OF are each 4 cm and perpendicular to the sides, so we find ar(△ABC) = ar(△OBC) + ar(△OCA) + ar(△OAB):

$$= \frac{1}{2} \times 4 \times 14 + \frac{1}{2} \times 4 \times (x + 6) + \frac{1}{2} \times 4 \times (x + 8)$$

$$= 2(2x + 28)$$

$$= 4(x + 14)$$

The two areas are equal, so:

$$4(x + 14) = \sqrt{48x(x + 14)}$$

$$\Rightarrow 16(x + 14)^2 = 48x(x + 14)$$

$$\Rightarrow x + 14 = 3x$$

$$\Rightarrow 2x = 14$$

$$\Rightarrow x = 7$$

(We divided both sides by $16(x + 14)$, which is not zero.)

So:
AB = 7 + 8 = 15
AC = 7 + 6 = 13

∴ AB = 15 cm and AC = 13 cm.

Let ABCD be a quadrilateral circumscribing a circle with centre O, touching AB, BC, CD and DA at P, Q, R and S respectively. Join OP, OQ, OR and OS.

In △OAP and △OAS:
OP = OS (radii)
AP = AS (tangents from the external point A)
OA = OA (common)

So by SSS congruence, △OAP ≅ △OAS, and:

$$\angle AOP = \angle AOS$$

Let this angle be $a$. In the same way, using the tangents from B, C and D:
$\angle BOP = \angle BOQ = b$
$\angle COQ = \angle COR = c$
$\angle DOR = \angle DOS = d$

The angles around O add up to 360°, so:

$$2a + 2b + 2c + 2d = 360°$$

$$\Rightarrow a + b + c + d = 180° \quad \cdots (1)$$

Now:
$\angle AOB = \angle AOP + \angle BOP = a + b$
$\angle COD = \angle COR + \angle DOR = c + d$

Using equation (1):

$$\angle AOB + \angle COD = a + b + c + d$$

$$\Rightarrow \angle AOB + \angle COD = 180°$$

In the same way:
$\angle BOC = b + c$
$\angle AOD = a + d$

$$\angle BOC + \angle AOD = a + b + c + d$$

$$\Rightarrow \angle BOC + \angle AOD = 180°$$

∴ Opposite sides subtend supplementary angles at the centre. Hence proved. $\blacksquare$