Class 10-NCERT Solutions-Chapter-10-Circles

Answer: A circle can have infinitely many tangents. There is a tangent at every point on the circle, and a circle consists of infinite points.

Answers:

(i) one

(ii) secant

(iii) two

(iv) point of contact

Solution:

Since tangent is perpendicular to the radius at the point of contact, \( \angle OPQ = 90^{\circ} \).

In right \( \Delta OPQ \):

\[ OQ^2 = OP^2 + PQ^2 \]

\[ 12^2 = 5^2 + PQ^2 \]

\[ 144 = 25 + PQ^2 \]

\[ PQ^2 = 144 - 25 = 119 \]

\[ PQ = \sqrt{119} \text{ cm} \]

Answer: (D) \( \sqrt{119} \) cm

Solution:

• Draw a circle with center O.
• Draw any straight line \( AB \).
• Draw a line \( CD \) parallel to \( AB \) intersecting the circle at two distinct points (Secant).
• Draw a line \( EF \) parallel to \( AB \) touching the circle at exactly one point (Tangent).

Solution:

Let O be the centre and T be the point of contact. \( \angle OTQ = 90^{\circ} \).

Given \( OQ = 25 \) cm, \( TQ = 24 \) cm. Radius \( OT = r \).

\[ OQ^2 = OT^2 + TQ^2 \]

\[ 25^2 = r^2 + 24^2 \]

\[ 625 = r^2 + 576 \]

\[ r^2 = 625 - 576 = 49 \]

\[ r = 7 \text{ cm} \]

Answer: (A) 7 cm

Solution:

Tangents are perpendicular to radii at points of contact. So, \( \angle OPT = 90^{\circ} \) and \( \angle OQT = 90^{\circ} \).

In quadrilateral POQT, the sum of angles is \( 360^{\circ} \).

\[ \angle PTQ + \angle OPT + \angle POQ + \angle OQT = 360^{\circ} \]

\[ \angle PTQ + 90^{\circ} + 110^{\circ} + 90^{\circ} = 360^{\circ} \]

\[ \angle PTQ + 290^{\circ} = 360^{\circ} \]

\[ \angle PTQ = 360^{\circ} - 290^{\circ} = 70^{\circ} \]

Answer: (B) 70°

Solution:

Given \( \angle APB = 80^{\circ} \).

In quadrilateral OAPB, \( \angle OAP = 90^{\circ}, \angle OBP = 90^{\circ} \).

So, \( \angle AOB = 360^{\circ} - (90 + 90 + 80) = 100^{\circ} \).

Triangles \( \Delta OAP \) and \( \Delta OBP \) are congruent (RHS). Therefore, \( \angle POA = \angle POB \).

\[ \angle POA = \frac{1}{2} \angle AOB = \frac{1}{2} \times 100^{\circ} = 50^{\circ} \]

Answer: (A) 50°

Solution:

Let PQ be a diameter of the circle with centre O.

Let AB be the tangent at P and CD be the tangent at Q.

Since radius is perpendicular to tangent, \( OP \perp AB \) and \( OQ \perp CD \).

Thus, \( \angle OPA = 90^{\circ} \) and \( \angle OQD = 90^{\circ} \).

Since PQ is a straight line, \( \angle OPA \) and \( \angle OQD \) are alternate interior angles (if we consider PQ as transversal).

Since alternate interior angles are equal (\( 90^{\circ} = 90^{\circ} \)), line AB is parallel to line CD.

Solution:

Let AB be a tangent to the circle at point P. Let O be the centre.

We know that the radius is perpendicular to the tangent at the point of contact. So, \( OP \perp AB \).

This means the line perpendicular to the tangent at P is the line containing the radius OP.

Since the radius OP passes through the centre O, the perpendicular at the point of contact must pass through the centre.

(Alternatively, assume it passes through another point O'. Then \( \angle O'PB = 90^{\circ} \) and \( \angle OPB = 90^{\circ} \), which implies O and O' lie on the same line perpendicular to AB at P).

Solution:

Let O be the centre and T be the point of contact of the tangent.

\( OA = 5 \) cm (Distance from centre).

\( AT = 4 \) cm (Length of tangent).

In right \( \Delta OTA \):

\[ OA^2 = OT^2 + AT^2 \]

\[ 5^2 = r^2 + 4^2 \]

\[ 25 = r^2 + 16 \]

\[ r^2 = 9 \Rightarrow r = 3 \text{ cm} \]

Answer: Radius is 3 cm.

Solution:

Let O be the common centre. Let AB be the chord of the larger circle touching the smaller circle at P.

\( OP \) is the radius of the smaller circle = 3 cm.

\( OA \) is the radius of the larger circle = 5 cm.

Since AB is tangent to the smaller circle, \( OP \perp AB \).

Perpendicular from centre to chord bisects the chord, so \( AP = PB \).

In right \( \Delta OPA \):

\[ OA^2 = OP^2 + AP^2 \]

\[ 5^2 = 3^2 + AP^2 \]

\[ 25 = 9 + AP^2 \Rightarrow AP^2 = 16 \Rightarrow AP = 4 \text{ cm} \]

Length of chord \( AB = 2 \times AP = 2 \times 4 = 8 \text{ cm} \).

Solution:

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

Using the property that tangents from an external point are equal in length:

• \( AP = AS \) (Tangents from A)
• \( BP = BQ \) (Tangents from B)
• \( CR = CQ \) (Tangents from C)
• \( DR = DS \) (Tangents from D)

Add LHS and RHS:

\[ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) \]

Since \( AP+BP=AB \), \( CR+DR=CD \), etc.,

\[ AB + CD = AD + BC \]

Hence proved.

Solution:

Join O to C (point of contact).

In \( \Delta OPA \) and \( \Delta OCA \):

• \( OP = OC \) (Radii)
• \( AP = AC \) (Tangents from A)
• \( AO = AO \) (Common)

Therefore, \( \Delta OPA \cong \Delta OCA \) (SSS). Thus \( \angle POA = \angle COA \). Let this be \( x \).

Similarly, \( \Delta OQB \cong \Delta OCB \). Thus \( \angle QOB = \angle COB \). Let this be \( y \).

POQ is a diameter (straight line) because tangents at P and Q are parallel.

So, \( \angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ} \).

\[ 2x + 2y = 180^{\circ} \]

\[ 2(x + y) = 180^{\circ} \Rightarrow x + y = 90^{\circ} \]

Since \( \angle AOB = x + y \), therefore \( \angle AOB = 90^{\circ} \).

Solution:

Let PA and PB be tangents from P touching circle at A and B. O is the centre.

We need to prove \( \angle APB + \angle AOB = 180^{\circ} \).

In quadrilateral OAPB:

• \( \angle OAP = 90^{\circ} \) (Radius perpendicular to tangent)
• \( \angle OBP = 90^{\circ} \)

Sum of angles in quadrilateral = \( 360^{\circ} \).

\[ \angle AOB + \angle OBP + \angle APB + \angle OAP = 360^{\circ} \]

\[ \angle AOB + 90^{\circ} + \angle APB + 90^{\circ} = 360^{\circ} \]

\[ \angle AOB + \angle APB = 360^{\circ} - 180^{\circ} = 180^{\circ} \]

Hence, the angles are supplementary.

Solution:

Let ABCD be a parallelogram circumscribing a circle.

Using the result from Q8 (Quadrilateral circumscribing a circle):

\[ AB + CD = AD + BC \]

Since ABCD is a parallelogram, opposite sides are equal: \( AB = CD \) and \( AD = BC \).

Substitute these into the equation:

\[ AB + AB = AD + AD \]

\[ 2AB = 2AD \Rightarrow AB = AD \]

Since adjacent sides are equal in a parallelogram, all sides are equal (\( AB = BC = CD = DA \)).

Therefore, ABCD is a rhombus.

Solution:

Let points of contact on BC, AC, AB be D, E, F respectively.

Given \( CD = 6 \) cm, \( BD = 8 \) cm. Radius \( r = 4 \) cm.

Using tangent properties:

• \( CE = CD = 6 \) cm
• \( BF = BD = 8 \) cm
• Let \( AE = AF = x \) cm

Sides of triangle:

• \( a = BC = 6 + 8 = 14 \) cm
• \( b = AC = 6 + x \) cm
• \( c = AB = 8 + x \) cm

Step 1: Area using Heron's Formula

Semi-perimeter \( s = \frac{14 + (6+x) + (8+x)}{2} = \frac{28+2x}{2} = 14 + x \).

Area = \( \sqrt{s(s-a)(s-b)(s-c)} \)

\( = \sqrt{(14+x)(x)(8)(6)} = \sqrt{48x(14+x)} \)

Step 2: Area using Triangles OBC, OCA, OAB

Area = Area(\( \Delta OBC \)) + Area(\( \Delta OCA \)) + Area(\( \Delta OAB \))

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

\( = 2(14 + 6 + x + 8 + x) = 2(28 + 2x) = 4(14 + x) \)

Step 3: Equate Areas

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

Squaring both sides:

\[ 48x(14+x) = 16(14+x)^2 \]

Divide by \( 16(14+x) \):

\[ 3x = 14 + x \]

\[ 2x = 14 \Rightarrow x = 7 \text{ cm} \]

Answer: \( AB = 8 + 7 = 15 \) cm, \( AC = 6 + 7 = 13 \) cm.

Solution:

Let ABCD circumscribe the circle with centre O. Let P, Q, R, S be points of contact on AB, BC, CD, DA.

Join radii OP, OQ, OR, OS. Join OA, OB, OC, OD.

In \( \Delta OAP \) and \( \Delta OAS \), \( AP = AS \) (tangents) and radii are equal. So \( \Delta OAP \cong \Delta OAS \).

Thus angles at centre are equal. Let these be \( \angle 1 \) and \( \angle 8 \). Similarly, adjacent angle pairs are equal (\( \angle 2=\angle 3, \angle 4=\angle 5, \angle 6=\angle 7 \)).

Sum of angles at centre = \( 360^{\circ} \).

\[ 2(\angle 1 + \angle 2 + \angle 5 + \angle 6) = 360^{\circ} \]

\[ (\angle 1 + \angle 2) + (\angle 5 + \angle 6) = 180^{\circ} \]

Notice \( \angle 1 + \angle 2 = \angle AOB \) and \( \angle 5 + \angle 6 = \angle COD \).

Therefore, \( \angle AOB + \angle COD = 180^{\circ} \).

Hence proved.