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

Given: A circle with centre O, tangent PQ at point of contact P.

To Prove: OP ⊥ PQ

Proof: Let Q be any point on tangent PQ other than P. Since tangent touches circle only at P, Q lies outside the circle.

Therefore $OQ > OP$ (radius is shortest distance from centre to any point; all other points of tangent are outside circle).

This is true for every point Q on PQ except P.

Hence OP is the shortest distance from O to line PQ.

Since perpendicular is the shortest distance from a point to a line:

OP ⊥ PQ. $\blacksquare$

Standard Q2 from NCERT Ex 10.2:

In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to: (A) 60°, (B) 70°, (C) 80°, (D) 90°

$\angle POQ + \angle PTQ = 180°$ (since OP ⊥ TP and OQ ⊥ TQ, so OPTQ is a cyclic quadrilateral with ∠P = ∠Q = 90°, sum = 360°)

$110° + \angle PTQ = 180°$… Wait:

$\angle OPT = \angle OQT = 90°$

In quadrilateral OPTQ: $\angle POQ + \angle OPT + \angle PTQ + \angle OQT = 360°$ $110° + 90° + \angle PTQ + 90° = 360°$ $\angle PTQ = 70°$

Answer: (B) 70°

In △OAP and △OBP: OA = OB (radii), PA = PB (tangent lengths), OP = OP (common).

By SSS: △OAP ≅ △OBP

So $\angle APO = \angle BPO = 40°$ (since $\angle APB = 80°$)

Also $\angle OAP = 90°$ (radius ⊥ tangent)

In △OAP: $\angle AOP + \angle OAP + \angle APO = 180°$ $\angle AOP + 90° + 40° = 180°$ $\angle AOP = 50°$

Answer: (A) 50°

Given: AB is a diameter of circle with centre O. Tangent PQ at A and tangent RS at B.

To prove: PQ ∥ RS

Proof:

Since PQ is tangent at A: OA ⊥ PQ → $\angle OAP = \angle OAQ = 90°$

Since RS is tangent at B: OB ⊥ RS → $\angle OBR = \angle OBS = 90°$

Since $\angle OAQ = \angle OBR = 90°$, these are alternate interior angles formed by transversal AB with lines PQ and RS.

Therefore PQ ∥ RS. $\blacksquare$

Given: Tangent PQ at point A of circle with centre O.

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

Proof by contradiction:

Suppose the perpendicular to PQ at A does not pass through O. Then let there be a point O’ (≠ O) on this perpendicular.

So O’A ⊥ PQ.

But we know OA ⊥ PQ (radius ⊥ tangent at point of contact).

This means both OA and O’A are perpendicular to PQ at the same point A, which is impossible (only one perpendicular can be drawn to a line at a given point).

This is a contradiction. Therefore, our assumption is wrong.

Hence, the perpendicular to PQ at A passes through O (the centre). $\blacksquare$

Let r = radius, OA = 5 cm, tangent length AT = 4 cm.

OT ⊥ AT (radius ⊥ tangent), so in right △OTA:

∴ Radius = 3 cm.

Let O be the common centre, AB be a chord of the larger circle (radius 5 cm) that is tangent to the smaller circle (radius 3 cm) at M.

OM ⊥ AB (tangent ⊥ radius), so OM = 3 cm.

In right △OMA:

$AB = 2 \times AM = 8$ cm (perpendicular from centre bisects chord).

∴ Length of chord = 8 cm.

Since tangents from an external point are equal:

From A: $AP = AS$ … (1) From B: $BP = BQ$ … (2) From C: $CQ = CR$ … (3) From D: $DR = DS$ … (4)

(where P, Q, R, S are points of tangency on AB, BC, CD, DA respectively)

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

Hence proved. $\blacksquare$

Since tangents from external point A: $AP = AC$ (P is point of contact on XY)

So △OAP ≅ △OAC (OA common, AP = AC, OP = OC = radius) → $\angle OAP = \angle OAC$, i.e., OA bisects $\angle PAC$.

Similarly, OB bisects $\angle QBC$ (Q is point of contact on X’Y’).

Since XY ∥ X’Y’, PA ∥ QB: $\angle PAC + \angle QBC = 180°$ (co-interior angles with transversal AB)

$\angle OAC + \angle OBC = 90°$

In △AOB: $\angle AOB = 180° - (\angle OAC + \angle OBC) = 180° - 90° = 90°$

Hence $\angle AOB = 90°$. $\blacksquare$

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

$\angle OAP = \angle OBP = 90°$ (radius ⊥ tangent)

In quadrilateral OAPB:

Therefore, the angle between the tangents ($\angle APB$) and the angle subtended at the centre ($\angle AOB$) are supplementary. $\blacksquare$

Let ABCD be a parallelogram circumscribing a circle with centre O.

From Q8: $AB + CD = AD + BC$ … (1)

Since ABCD is a parallelogram: $AB = CD$ and $AD = BC$ … (2)

From (1): $2AB = 2AD \Rightarrow AB = AD$

Since all sides of parallelogram ABCD are equal ($AB = BC = CD = DA$), ABCD is a rhombus. $\blacksquare$

Let BD = 8 cm, DC = 6 cm, radius = 4 cm.

Using tangent lengths from each vertex:

• From B: BD = BF = 8 cm (F is contact point on AB)
• From C: CD = CE = 6 cm (E is contact point on AC)
• From A: AF = AE = x (say)

So: $AB = AF + FB = x + 8$, $AC = AE + EC = x + 6$, $BC = 8 + 6 = 14$ cm.

Area of △ABC using Heron’s formula: $s = \dfrac{AB+BC+CA}{2} = \dfrac{(x+8)+14+(x+6)}{2} = x + 14$

$\text{ar}(\triangle ABC) = \sqrt{s(s-a)(s-b)(s-c)}$

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

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

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

Also, area = $r \times s = 4(x+14)$

∴ $AB = x + 8 = 15$ cm, $AC = x + 6 = 13$ cm.

Let ABCD be a quadrilateral circumscribing a circle with centre O. Let the circle touch AB, BC, CD, DA at P, Q, R, S respectively.

Join OP, OQ, OR, OS.

In △OAP and △OAS: $AP = AS$ (tangents from A), $OP = OS = r$, OA = OA. By SSS: △OAP ≅ △OAS → $\angle AOP = \angle AOS$

Let $\angle AOP = \angle AOS = \alpha$

Similarly: $\angle BOP = \angle BOQ = \beta$, $\angle COQ = \angle COR = \gamma$, $\angle DOR = \angle DOS = \delta$

Total angles around O:

$\angle AOB = \angle AOP + \angle BOP = \alpha + \beta$

$\angle COD = \angle COR + \angle DOR = \gamma + \delta$

$\angle AOB + \angle COD = \alpha + \beta + \gamma + \delta = 180°$

Similarly: $\angle BOC + \angle DOA = 180°$

Hence, opposite sides subtend supplementary angles at the centre. $\blacksquare$