Circles — NCERT Solutions

Given:

• Distance of point A from centre O: $OA = 5$ cm
• Length of tangent from A: $AB = 4$ cm
• Let radius $= OB = r$

Concept: The tangent is perpendicular to the radius at the point of contact.

So $\angle OBA = 90°$.

Applying Pythagoras Theorem in right $\triangle OBA$:
$$OA^2 = OB^2 + AB^2$$
$$5^2 = r^2 + 4^2$$
$$25 = r^2 + 16$$
$$r^2 = 25 - 16 = 9$$
$$\boxed{r = 3 \text{ cm}}$$

The radius of the circle is 3 cm.

Given:

• Radius of larger circle $= R = 5$ cm
• Radius of smaller circle $= r = 3$ cm
• Let AB be a chord of the larger circle that is tangent to the smaller circle.

Let O be the common centre, and let the chord AB touch the smaller circle at point P.

Concept: The radius to the point of tangency is perpendicular to the tangent.

So $OP \perp AB$, which means $\angle OPA = 90°$.

Also, since the perpendicular from the centre bisects the chord:
$$AP = PB$$

In right $\triangle OPA$:
$$OA^2 = OP^2 + AP^2$$
$$5^2 = 3^2 + AP^2$$
$$25 = 9 + AP^2$$
$$AP^2 = 16$$
$$AP = 4 \text{ cm}$$

Length of chord:
$$AB = 2 \times AP = 2 \times 4 = \boxed{8 \text{ cm}}$$

Given: Quadrilateral ABCD circumscribes a circle. The circle touches sides AB, BC, CD and DA at points P, Q, R and S respectively.

To Prove: $AB + CD = AD + BC$

Proof:

We know that tangent segments drawn from an external point to a circle are equal in length.

From vertex A: $AP = AS$ ...(1)

From vertex B: $BP = BQ$ ...(2)

From vertex C: $CQ = CR$ ...(3)

From vertex D: $DR = DS$ ...(4)

Adding (1), (2), (3) and (4):
$$AP + BP + CQ + DR = AS + BQ + CR + DS$$

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

$$AB + CD = AD + BC$$

Hence proved.

Given:

• XY and X′Y′ are two parallel tangents to a circle with centre O, touching at P and Q respectively.
• AB is a third tangent touching the circle at C, meeting XY at A and X′Y′ at B.

To Prove: $\angle AOB = 90°$

Proof:

Join OA, OB, and OC.

In $\triangle OAP$ and $\triangle OAC$:

• $AP = AC$ (tangents from external point A are equal)
• $OA = OA$ (common)
• $OP = OC$ (radii)

By SSS congruence: $\triangle OAP \cong \triangle OAC$

Therefore: $\angle POA = \angle COA$

So OA bisects $\angle POC$:
$$\angle POA = \angle COA = \frac{1}{2}\angle POC \quad \text{...(1)}$$

Similarly, in $\triangle OBQ$ and $\triangle OBC$:

• $BQ = BC$ (tangents from external point B are equal)
• $OB = OB$ (common)
• $OQ = OC$ (radii)

By SSS congruence: $\triangle OBQ \cong \triangle OBC$

Therefore: $\angle QOB = \angle COB$

So OB bisects $\angle QOC$:
$$\angle QOB = \angle COB = \frac{1}{2}\angle QOC \quad \text{...(2)}$$

Now, since XY $\parallel$ X′Y′ and PQ is a transversal (a diameter):
$$\angle POQ = 180° \quad \text{(P and Q are ends of a diameter, as XY ∥ X'Y')}$$

From (1) and (2):
$$\angle AOB = \angle COA + \angle COB = \frac{1}{2}\angle POC + \frac{1}{2}\angle QOC$$
$$= \frac{1}{2}(\angle POC + \angle QOC) = \frac{1}{2} \times \angle POQ = \frac{1}{2} \times 180° = 90°$$

$$\boxed{\angle AOB = 90°}$$

Hence proved.

Given: PA and PB are two tangents from external point P to a circle with centre O. A and B are the points of contact.

To Prove: $\angle APB + \angle AOB = 180°$

Proof:

Since PA is a tangent at A:
$$OA \perp PA \implies \angle OAP = 90° \quad \text{...(1)}$$

Since PB is a tangent at B:
$$OB \perp PB \implies \angle OBP = 90° \quad \text{...(2)}$$

In quadrilateral OAPB, the sum of all angles $= 360°$:
$$\angle OAP + \angle APB + \angle OBP + \angle AOB = 360°$$

Substituting from (1) and (2):
$$90° + \angle APB + 90° + \angle AOB = 360°$$
$$\angle APB + \angle AOB = 360° - 180°$$
$$\boxed{\angle APB + \angle AOB = 180°}$$

Hence, the angle between the two tangents is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Hence proved.

Given: A parallelogram ABCD circumscribes a circle with centre O. The circle touches sides AB, BC, CD and DA at points P, Q, R and S respectively.

To Prove: ABCD is a rhombus, i.e., $AB = BC = CD = DA$.

Proof:

We know that tangent segments from an external point are equal.

From vertex A: $AP = AS$ ...(1)

From vertex B: $BP = BQ$ ...(2)

From vertex C: $CQ = CR$ ...(3)

From vertex D: $DR = DS$ ...(4)

Adding all four:
$$(AP + BP) + (CQ + DR) = (AS + DS) + (BQ + CR)$$
$$AB + CD = AD + BC \quad \text{...(5)}$$

Since ABCD is a parallelogram:
$$AB = CD \quad \text{and} \quad AD = BC \quad \text{...(6)}$$

Substituting (6) into (5):
$$AB + AB = AD + AD$$
$$2AB = 2AD$$
$$AB = AD$$

Since $AB = CD$ and $AD = BC$ (opposite sides of parallelogram), and $AB = AD$:
$$AB = BC = CD = DA$$

Therefore, ABCD is a rhombus.

Hence proved.

Given:

• Circle of radius 4 cm inscribed in $\triangle ABC$
• $BD = 8$ cm, $DC = 6$ cm
• Let the circle touch AB at E and AC at F

Using the property that tangent segments from an external point are equal:

From B: $BE = BD = 8$ cm ...(1)

From C: $CF = CD = 6$ cm ...(2)

From A: $AE = AF = x$ (say) ...(3)

So the sides are:
$$BC = BD + DC = 8 + 6 = 14 \text{ cm}$$
$$AB = AE + EB = x + 8$$
$$AC = AF + FC = x + 6$$

Finding x using the area method:

Let $s$ be the semi-perimeter:
$$s = \frac{AB + BC + CA}{2} = \frac{(x+8) + 14 + (x+6)}{2} = \frac{2x + 28}{2} = x + 14$$

Area of $\triangle ABC$ using Heron's formula:
$$s - AB = (x+14) - (x+8) = 6$$
$$s - BC = (x+14) - 14 = x$$
$$s - CA = (x+14) - (x+6) = 8$$

$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)(6)(x)(8)} = \sqrt{48x(x+14)}$$

Also, Area $=$ inradius $\times$ semi-perimeter:
$$\text{Area} = r \times s = 4 \times (x+14)$$

Equating both expressions:
$$4(x+14) = \sqrt{48x(x+14)}$$

Squaring both sides:
$$16(x+14)^2 = 48x(x+14)$$
$$16(x+14) = 48x$$
$$x + 14 = 3x$$
$$14 = 2x$$
$$x = 7 \text{ cm}$$

Therefore:
$$AB = x + 8 = 7 + 8 = \boxed{15 \text{ cm}}$$
$$AC = x + 6 = 7 + 6 = \boxed{13 \text{ cm}}$$

Given: A quadrilateral ABCD circumscribes a circle with centre O. The circle touches AB, BC, CD and DA at P, Q, R and S respectively.

To Prove:
$$\angle AOB + \angle COD = 180° \quad \text{and} \quad \angle AOD + \angle BOC = 180°$$

Proof:

Join OP, OQ, OR and OS.

In $\triangle OAP$ and $\triangle OAS$:

• $AP = AS$ (tangents from A)
• $OP = OS$ (radii)
• $OA = OA$ (common)

By SSS: $\triangle OAP \cong \triangle OAS$

$$\Rightarrow \angle AOP = \angle AOS = \angle 1 \text{ (say)}$$

Similarly:

• $\angle BOP = \angle BOQ = \angle 2$
• $\angle COQ = \angle COR = \angle 3$
• $\angle DOR = \angle DOS = \angle 4$

Since the sum of all angles around O is $360°$:
$$2\angle 1 + 2\angle 2 + 2\angle 3 + 2\angle 4 = 360°$$
$$\angle 1 + \angle 2 + \angle 3 + \angle 4 = 180° \quad \text{...(i)}$$

Now:
$$\angle AOB + \angle COD = (\angle 1 + \angle 2) + (\angle 3 + \angle 4)$$
$$= (\angle 1 + \angle 2 + \angle 3 + \angle 4) = 180° \quad \text{[from (i)]}$$

Also:
$$\angle AOD + \angle BOC = (\angle 1 + \angle 4) + (\angle 2 + \angle 3)$$
$$= (\angle 1 + \angle 2 + \angle 3 + \angle 4) = 180° \quad \text{[from (i)]}$$

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.

Hence proved.