**GSEB Solutions for Class 9 Mathematics – Circle (English Medium)**

GSEB SolutionsMathsScience

**Exercise – 12**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

**Solution 8:**

**Solution 9:**

**Solution 10****:**

**Solution 11:**

**Solution 12(1):**

a. in the interior of the circle

**Solution 12(2):**

a. in the interior of the circle

**Solution 12(3):**

d. a chord joining the end-points of the semicircle arc.

**Solution 12(4):**

d. a radius

**Solution 12(5):**

c. an equilateral triangle

For ⨀(O, x), m∠ AOB = 60°

In ∆ OAB,

Radius = OA = OB = x

∴ ∠AOB + ∠OBA + ∠OAB = 180°

∴ 60° + x + x = 180°

∴ x = 60°

∴ ∆ OAB is a triangle.

**Solution 12(6):**

d. isosceles triangle

In ∆ OAB,

OA = OB = Radius

Thus, Δ OAB is isosceles triangle.

**Solution 12(7):**

**Solution 12(8):**

c. 60°

**Solution 12(9):**

b. a minor sector

**Solution 12(10):**

b. the midpoint of the diameter

**Solution 12(11):**

**Solution 12(12):**

**Solution 12(13):**

b. one

**Solution 12(14):**

d. infinite

**Solution 12(15):**

c. the centre of the circle passing through all A, B, C.

**Solution 12(16):**

d. a line passing through one of the end points of the common chord.

**Solution 12(17):**

a. 130° and 50°

The measure of the remaining angles = 180° – 50° = 130° and 180° – 100° = 80°.

**Solution 12(18):**

**Solution 12(19):**

d. 35°

m∠ ADC = 180 – m∠ ABC = 180° – 100° = 80°

m∠ADC = m∠ADB + m∠BDC

∴ 80° = m∠ADB + 45°

∴m∠ADB = 35°

**Solution 12(20):**

c. 50°

∠ APB is inscribed in a semicircle.

In ∆ PAB,

m∠ APB + m∠ PAB + m∠ PBA = 180°

∴ 90° + 40° + m∠ PBA = 180°

∴ m∠ PBA = 180° – 90° – 40° = 50°

**Solution 12(21):**

d. 120°

In equilateral triangle ABC,

m∠ ACB = 60°

P is the circumcentre of equilateral ΔABC.

m∠ APB = 2 × m∠ ACB = 2 × 60° = 120°

**Exercise – 12.1**

**Solution 1:**

- If two circles having centres P and Q are concentric, then points P and Q represent the same point. ∴P = Q
- If two circles having centres P and Q are congruent, then their radii are equal.
- In a circle with centre O, if P is in the interior and Q is in the exterior, then OQ is larger than OP.

**Solution 2:**

**False**A line-sgement joining the centre to any point of the circle is the radius of the circle.**True**An arc with the diameter as its endpoints is always a semi-circle.**False**The set of points equidistant from a fixed point in a plane form a circle whereas the set of points equidistant from a fixed point in space form a sphere.**False**The union of any two radii of a circle is not a diameter of the circle. The union of two radii of a circle is a diameter of the circle only if the radii lie on the same line.

**Exercise – 12.2**

**Solution 1:**

**Exercise – 12.3**

**Solution 1:**

The possible number of points of intersection of two distinct circles in a plane depends on the distance between their centres and the sum and positive difference between their radii.

Consider ⨀ (A, R_{1}) and ⨀ (B, R_{2}) lying in a plane. Then the distance between their centres is AB and the sum of their radii is R_{1 }+ R_{2}.

1. If AB > R_{1 }+ R_{2}, then the circles do not intersect each other. The number of points of intersection of the two circles is zero.

2. If AB = R_{1 }+ R_{2}, the circles intersect each other at only one point.

3. If |R_{1 }– R_{2}| < AB < R_{1 }+ R_{2}, the circles intersect each other at two distinct points.

4. If AB = |R_{1 }– R_{2}|, the circles intersect each other at only one point.

5. If AB < | R_{1 }– R_{2}|, the circles do not intersect each other.

Thus, the possible number of points of intersection of two distinct circles in a plane is zero, one or two.

**Solution 2:**

**Exercise – 12.4**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Exercise – 12.5**

**Solution 1:**

The measure of the angle subtended by a minor arc of a circle at the centre is twice the measure of an angle subtended by the arc at any point on the remaining part of the circle.

m∠AOB = 2 × m∠ADB = 2 × 45° = 90°

**Solution 2:**

In ∆ABC,

m∠BAC + m∠ABC + m∠ACB = 180°

∴m∠BAC + 49° + 51° = 180°

∴m∠BAC + 100° = 180°

∴m∠BAC = 180° – 100°

∴m∠BAC = 80°

Also, ∠BAC and ∠BDC are angles in the same segment.

∴m∠BDC = m∠BAC

∴m∠BDC = 80°

**Solution 3:**

**Solution 4:**

**Solution 5:**

∠SQR and ∠SPR are angles in the same segment.

∴m∠SPR = m∠SQR = 70°

∴m∠SPQ = m∠SPR + m∠QPR (∵ R lies in the interior of ∠SPQ.)

∴m∠SPQ = 70° + 30° = 100°

In cyclic □ PQRS,

m∠QRS + m∠SPQ = 180°

∴ m∠QRS + 100° = 180°

∴ m∠QRS = 180° – 100° = 80°

In ∆PQR, PQ = PR

∴ m∠PRQ = m∠PQR

Now,

m∠PRQ + m∠PQR + m∠QPR = 180°

∴ m∠PQR + m∠PQR + 30° = 180° …[ m∠QPR = 30° ]

∴ 2m∠PQR = 150°

∴ m∠PQR = 75°

m∠PQR = m∠PQS + m∠SQR

∴ 75° = m∠PQS + 70° …[m∠PQR = 75° and m∠SQR = 70°]

∴ m∠PQS = 5°

∠PQS and ∠PRS are angles in the same segment.

∴m∠PRS = m∠PQS

∴m∠PRS = 5°

∴m∠ERS = 5° (∵ P – E – R)

**Solution 6:**

**Solution 7:**

**Solution 8:**

**Solution 9:**

**Solution 10:**