Circles

As we know that

the largest chord of a circle is its diameter.

Diameter, AB = 20 cm

Chord CD = half of length of chord AB = $\frac{1}{2} \times 18$

= 9 cm.

OP is radius of the circle. So, radius is half of the diameter.

OP = $\frac{1}{2} \times 18$ = 9 cm

∴ OP = 9 cm and CD = 9 cm

Hence, option 4 is the correct option.

The largest chord of a circle is its diameter.

∴ Diameter = 20 cm

As we know that radius is half of the diameter.

Radius = $\frac{1}{2} \times 20$ = 10 cm

Hence, option 3 is the correct option.

The central angle θ for the major segment is greater than 180°, leading to a larger sector area, and consequently, a larger segment area.

∴ Major segment is bigger than the semi-circle.

Hence, option 2 is the correct option.

Given, radius of a circle = 5 cm.

The largest chord of a circle is its diameter.

As we know that radius is half of the diameter.

⇒ Diameter = 2 x radius = 2 x 5 cm = 10 cm

∴ Length of its longest chord = 10 cm

Hence, option 3 is the correct option.

Given, diameter of a circle = 30 cm.

As we know that radius is half of the diameter.

⇒ Radius = $\frac{1}{2}$ x diameter= $\frac{1}{2}$ x 30

= 15 cm.

The distance from the center O to point P is 20 cm. Since OP = 20 cm is greater than the radius of the circle (15 cm).

∴ Point P lies outside the circle.

Hence, option 3 is the correct option.

The angle between a line AB and the diameter of a circle is 90°; line AB can be either a tangent of the circle or cannot be a tangent of the circle.

Thus, nothing can be confirmed.

Hence, option 3 is the correct option.

A segment of a circle is the area bounded by a chord of a circle and the area of the circle between the two endpoints of the chord.

A sector of a circle is a portion of a circle created by the arc of the circle and two radii.

∴ A sector of a circle can be equal to a segment of the same circle when the sector is a semicircle.

Yes, a sector of a circle can be equal to its segment.

Hence, option 1 is the correct option.

Given, length of arc AQC is double of arc APB.

If arc AQC is twice the length of arc APB, then the angle subtended by arc AQC at the center (∠AOC) will also be twice the angle subtended by arc APB (∠AOB).

∴ ∠AOC = 2 x ∠AOB.

⇒ ∠AOC = 2 x 65°

⇒ ∠AOC = 130°.

Hence, option 2 is the correct option.

A tangent line touches the circle at one point. A radius connects the center to the circumference.

The angle between a radius and tangent at the point of contact is always 90°. This is a fundamental property of circles.

∴ Both the statements are true.

Hence, option 1 is the correct option.

As we know that radius is half of the diameter.

Diameter = 2 x radius = 2 x 20 = 40 cm.

We know that,

Longest chord of a circle is equal to its diameter.

So, reason (R) is true.

Longest chord = 40 cm

So, assertion (A) is false.

∴ A is false, but R is true.

Hence, option 4 is the correct option.

Original radius = r cm

Original diameter = 2r cm

New radius = (r + 3) cm

New diameter = 2(r + 3) cm

Increase in length of diameter = New diameter - Original diameter

= 2(r + 3) - 2r

= 2r + 2 x 3 - 2r

= 2 x 3

= 6 cm.

∴ Both A and R are correct, and R is the correct explanation for A.

Hence, option 1 is the correct option.

Given, radius of a circle = 8 cm and the length of its minor arc = 8 cm

Circumference of the circle = 2πr = 2 x π x 8 = 16π

We know that,

Length of minor arc + length of major arc = Circumference of the circle

So, reason (R) is true.

⇒ 8 + length of major arc = 16π

⇒ length of major arc = 16π - 8

⇒ length of major arc = 8(2π - 1) cm

So, assertion (A) is true.

∴ Both A and R are correct, and R is the correct explanation for A.

Hence, option 1 is the correct option.

Given, AC = 4 cm and AB = 3 cm

Since BC is the diameter, angle BAC is a right angle.

Triangle ABC is a right triangle with hypotenuse BC.

Using Pythagoras theorem,

BC² = AB² + AC²

So, reason (R) is true.

⇒ BC² = 3² + 4²

⇒ BC² = 9 + 16

⇒ BC² = 25

⇒ BC = $\sqrt{25}$

⇒ BC = 5 cm.

So, assertion (A) is false.

∴ A is false, but R is true.

Hence, option 4 is the correct option.

(i) AB = Diameter

(ii) Radius = OE = OA = OB

(iii) Chords = CD and AB

(iv) Diameter = AB

(v) AB = 2 x OE = OA = OB

(i) Given, M is a fixed point. P moves such that PM = 10 cm.

A circle is a set of points equidistant from a center or a fixed point.

Hence, the figure is circle with M as center.

(ii) M is the center of the circle and PM is the radius of the circle.

Hence, the length of the radius of the circle = 10 cm.

(iii) As we know that the diameter is twice the radius.

Diameter = 2 x 10 cm = 20 cm

Hence, the length of the diameter of the circle = 20 cm.

The longest chord is the diameter.

The diameter is 20 cm.

A chord of length 16 cm can be drawn since 16 < 20.

Hence, a chord of length 16 cm can be drawn in this circle.

The radius of a circle = 6 cm

As, the diameter is twice the radius.

Diameter = 2 x 6 cm = 12 cm.

(i) OA = 4.8 cm

As, 4.8 < 6 (Radius)

Hence, A lies inside the circle.

(ii) OB = 7.5 cm

As, 7.5 > 6 (Radius)

Hence, B lies outside the circle.

(iii) OC = 6 cm

As, 6 = 6 (Radius)

Hence, C lies on the circumference of the circle.

(i) An arc is a part of the circumference of the circle.

(ii) Diameter of a circle bisects circumference of the circle.

(iii) The part of the circumference greater than the semicircle is called major-arc.

(iv) Sector of a circle is its region bounded by two radii and an arc.

(v) The segment of a circle is the region bounded by a chord and an arc.

(vi) A tangent of a circle meets the circle at one point only.

(vii) The number of tangents that can be drawn through a point on its circumference = only one.

(viii) The number of tangent that can be drawn through a point outside the circle is two.

Given, diameter = 30 cm

As we know that the diameter is twice the radius.

⇒ Radius = $\frac{\text{Diameter}}{2} = \frac{30}{2}$ = 15 cm.

(i) It is given that OP = 39 cm.

In ∆ OPA,

∠OAP = 90°.

By pythagoras theorem,

⇒ OA² + PA² = OP²

⇒ 15² + PA² = 39²

⇒ 225 + PA² = 1521

⇒ PA² = 1521 - 225

⇒ PA² = 1296

⇒ PA = $\sqrt{1296}$

⇒ PA = 36 cm.

Hence, the length of tangent PA = 36 cm.

(ii) Given,

PA = 20 cm

⇒ OA² + PA² = OP²

⇒ 15² + 20² = OP²

⇒ 225 + 400 = OP²

⇒ OP² = 625

⇒ OP = $\sqrt{625}$

⇒ OP = 25 cm.

Hence, the distance between O and P = 25 cm.

(i) AB is diameter of the circle.

Hence, angle APB is the angle of semi-circle.

(ii) As, the angle subtended by a diameter at any point on the circumference of a circle is a right angle (90°).

Hence, angle APB = 90°.

(iii) Given,

AP = 12 cm and OA = 10 cm

⇒ OA = OB (Radii of the circle)

⇒ OB = 10 cm

From figure,

⇒ AB = OA + OB = 10 cm + 10 cm = 20 cm

Since, angle APB = 90°.

Using pythagoras theorem,

⇒ AP² + PB² = AB²

⇒ 12² + PB² = 20²

⇒ 144 + PB² = 400

⇒ PB² = 400 - 144

⇒ PB² = 256

⇒ PB = $\sqrt{256}$

⇒ PB = 16 cm.

Hence, AB = 20 cm and PB = 16 cm.

(iv) Given,

⇒ 4AP = 3PB = 12 cm

⇒ AP = $\frac{12}{4}$ = 3 cm

⇒ PB = $\frac{12}{3}$ = 4 cm

Since, angle APB = 90°.

Using pythagoras theorem,

⇒ AP² + PB² = AB²

⇒ 3² + 4² = AB²

⇒ 9 + 16 = AB²

⇒ 25 = AB²

⇒ AB = $\sqrt{25}$

⇒ AB = 5 cm

Now, OA = OB (Radii of the circle)

From figure,

⇒ AB = OA + OB

⇒ AB = 2 x OA

⇒ 5 = 2 x OA

⇒ OA = $\frac{5}{2}$ = 2.5 cm.

Hence, the radius of the circle = 2.5 cm.

Consider a circle with center O. Let AP is the tangent of the circle.

We know that,

Tangent and radius to the circle at point of contact are perpendicular to each other.

In the ∆OPA,

∠OAP = 90°.

By pythagoras theorem,

⇒ OA² + PA² = OP²

⇒ 5² + PA² = 13²

⇒ 25 + PA² = 169

⇒ PA² = 169 - 25

⇒ PA² = 144

⇒ PA = $\sqrt{144}$

⇒ PA = 12 cm

Hence, the length of the tangent = 12 cm.

We know that,

Angle in a semi-circle is a right angle.

So, angle ACB = 90°.

In ∆ ACB,

Using angle sum property,

⇒ ∠ABC + ∠ACB + ∠CAB = 180°

⇒ ∠ABC + 90° + 35° = 180°

⇒ ∠ABC + 125° = 180°

⇒ ∠ABC = 180° - 125°

⇒ ∠ABC = 55°.

Hence, angle ABC = 55°.

Steps of construction :

1. Mark a point O.

2. With O as center and radius equal to 3 cm draw a circle.

3. Draw OA as the radius of the circle.

4. Draw another radius OB at an angle of 60° with OA.

5. Join A and B. Measure AB.

Hence, the length of AB = 3 cm.

Steps of construction:

1. Draw a circle of radius 4 cm and mark the centre as C.

2. Mark a point A on the circle.

3. Draw an arc with A as centre and radius 6 cm to cut the circle at B.

4. Join the line AB.

5. Shade the segment which contains the centre 'C' by vertical lines.

6. Shade the segment which does not contain the centre 'C' by horizontal lines.

Steps of construction:

1. Draw a circle of any radius and mark the centre as P.

2. Draw a straight line PB.

3. Draw an angle APB such that angle APB = 90°, where A is a point on the circumference of the circle.

4. Shade the major segment by vertical lines.

5. Shade the minor segment by horizontal lines.

Given, O is center of circle. AB is the diameter of the circle.

⇒ PA = PB (Given)

⇒ ∠PAB = ∠PBA = x (let) [Angles opposite to equal sides of a triangle are always equal]

We know that,

Angle in a semi-circle is a right angle.

⇒ ∠APB = 90°

In ΔAPB, using angle sum property

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\frac{90°}{2}$

⇒ x = 45°

⇒ ∠PAB = ∠PBA = 45°

Hence, all angles of the triangles are 45°, 45° and 90°.

We know that,

Angle in a semi-circle is a right angle.

⇒ ∠ACB = ∠ADB = 90°

In ΔACB, using angle sum property

⇒ ∠CAB + ∠CBA + ∠ACB = 180°

⇒ 35° + ∠CBA + 90° = 180°

⇒ 125° + ∠CBA = 180°

⇒ ∠CBA = 180° - 125°

⇒ ∠CBA = 55°

From figure,

⇒ ∠CBD = ∠CBA + ∠ABD = 55° + 55° = 110°.

Similarly, in ΔADB, using angle sum property

⇒ ∠DAB + ∠DBA + ∠ADB = 180°

⇒ ∠DAB + 55° + 90° = 180°

⇒ ∠DAB + 145° = 180°

⇒ ∠DAB = 180° - 145°

⇒ ∠DAB = 35°

From figure,

⇒ ∠CAD = ∠CAB + ∠DAB = 35° + 35° = 70°.

Hence, angles CAD = 70° and angle CBD = 110°.

Given, PQ is diameter of the circle.

PR = QR and PR = 4 cm

⇒ QR = 4 cm

As we know that,

Angle in a semi-circe is a right angle.

⇒ ∠PRQ = 90°

Triangle PQR is a right triangle with hypotenuse PQ. Using Pythagoras theorem,

⇒ PQ² = PR² + RQ²

⇒ PQ² = 4² + 4²

⇒ PQ² = 16 + 16

⇒ PQ² = 32

⇒ PQ = $\sqrt{32}$ = 5.66 cm

Radius (OP) = $\frac{1}{2} \times PQ = \frac{1}{2} \times 5.66$

= 2.83 cm.

Hence, the radius of the given circle = 2.83 cm.

(i) PQ and PR are the tangent to the circle with centre O.

As we know that,

Through any point, outside a circle at most two tangent can be drawn and both of these tangent are always equal to each other.

Here, point P lies outside the circle and the two tangent PQ and PR are drawn to the circle.

Hence, PQ = PR.

(ii) When point A lies outside the circle and the two tangent AQ and AB are drawn to the circle. Then,

AQ = AB ....................(1)

From figure,

⇒ PQ = PA + AQ

From equation (1),

⇒ PQ = PA + AB

Hence, proved that PQ = PA + AB.

(iii) When point C lies outside the circle and the two tangent CR and CB are drawn to the circle. Then

CR = CB ....................(2)

From figure,

⇒ PR = PC + CR

From equation (2),

⇒ PR = PC + CB

Hence, proved that PR = PC + CB.

(iv) Perimeter of triangle PAC = PA + PC + AC

= PA + PC + (AB + CB)

= PA + PC + (AQ + CR)

= (PA + AQ) + (PC + CR)

= PQ + PR

Hence, proved that PQ + PR = Perimeter of Δ PAC.

O is center of the circle. PA is the tangent of the circle.

We know that,

Tangent and radius through the center at point of contact are perpendicular to each other.

(i) Given, OP = 20 cm and PA = 16 cm

In the ∆OPA as we know that ∠OAP = 90°.

As we know Pythagoras theorem can be used in a right angled triangle,

⇒ OA² + PA² = OP²

⇒ OA² + 16² = 20²

⇒ OA² + 256 = 400

⇒ OA² = 400 - 256

⇒ OA² = 144

⇒ OA = $\sqrt{144}$

⇒ OA = 12 cm

By formula,

Diameter = 2 x radius = 2 x 12 = 24 cm.

Hence, the diameter of the circle = 24 cm.

(ii) Given, diameter = 20 cm and PA = 24 cm

Radius (OA) = $\frac{1}{2} \times 20$ = 10 cm

In the ∆ OPA as we know that ∠OAP = 90°.

As we know Pythagoras theorem can be used in a right angled triangle,

⇒ OA² + PA² = OP²

⇒ 10² + 24² = OP²

⇒ 100 + 576 = OP²

⇒ OP² = 676

⇒ OP = $\sqrt{676}$

⇒ OP = 26 cm.

Hence, the length of OP = 26 cm.

PA = PB (Given)

⇒ ∠PAB = ∠PBA = x (let) [Angles opposite to equal sides of a triangle are always equal]

As we know that the angle subtended by a diameter at any point on the circumference of a circle is a right angle (90°).

⇒ ∠APB = 90°

In ΔAPB, using angle sum property

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\frac{90°}{2}$

⇒ x = 45°

⇒ ∠PAB = ∠PBA = 45°

Hence, the value of x = 45°.

As we know that the angle subtended by a diameter at any point on the circumference of a circle is a right angle (90°).

⇒ ∠APB = 90°

According to exterior angle theorem,

Exterior angle of a triangle is equal to the sum of the two opposite interior angles.

⇒ ex. ∠A = ∠APB + ∠ABP

⇒ x = 90° + 60° = 150°

Hence, the value of x = 150°.

As we know that the angle subtended by a diameter at any point on the circumference of a circle is a right angle (90°).

⇒ ∠APB = 90°

According to exterior angle theorem,

Exterior angle of a triangle is equal to the sum of the two opposite interior angles.

⇒ ex. ∠A = ∠APB + ∠ABP

⇒ 130° = 90° + x

⇒ x = 130° - 90°

⇒ x = 40°.

Hence, the value of x = 40°.

At the point of contact, the angle between tangent and the radius is always 90°.

⇒ ∠APB = 90°

In ΔAPB, using angle sum property

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ x + 52° + 90° = 180°

⇒ x + 142° = 180°

⇒ x = 180° - 142°

⇒ x = 38°.

Hence, the value of x = 38°.

At the point of contact, the angle between tangent and the radius is always 90°.

⇒ ∠OAB = 90°

Triangle OAB is a right triangle with hypotenuse OB. Using Pythagoras theorem,

⇒ OB² = OA² + AB²

⇒ 10² = 8² + AB²

⇒ 100 = 64 + AB²

⇒ AB² = 100 - 64

⇒ AB² = 36

⇒ AB = $\sqrt{36}$

⇒ AB = 6 cm

We know that,

Tangents from an exterior point to a circle are equal in length.

⇒ BC = BA

⇒ x = 6 cm.

Hence, the value of x = 6 cm.

Angle in semi-circle is a right angle.

⇒ ∠APB = 90°

Triangle APB is a right triangle with hypotenuse AB. Using Pythagoras theorem,

⇒ AB² = PA² + PB²

⇒ AB² = 10² + 24²

⇒ AB² = 100 + 576

⇒ AB² = 676

⇒ AB = $\sqrt{676}$

⇒ AB = 26 cm

Radius is half of the diameter.

⇒ x = $\frac{AB}{2}$

⇒ x = $\frac{1}{2} \times 26$ = 13 cm.

Hence, the value of x = 13 cm.