ML Aggarwal Class 8 Maths Solutions Chapter 15 Circles

Question 1. Draw a circle with centre O and radius 2.5 cm. Draw two radii OA and OB such that ∠AOB = 60°. Measure the length of the chord AB.
Answer: Follow these steps to complete this construction: First, create a circle using O as the centre with a radius measuring 2.5 cm. Next, mark any point A on the circle and connect it to O with a line segment. Then, at centre O, draw an angle of 60° and mark the second ray meeting the circle at point B. Finally, join points A and B with a straight line. When you measure the length of chord AB, you will get approximately 2.5 cm.
In simple words: Draw a circle. Make two radii at 60 degrees to each other. The line joining their endpoints is the chord.

Exam Tip: Use a protractor to measure the 60° angle accurately at the centre. The chord length can be verified using the property that when the central angle equals 60°, the chord equals the radius.

Question 2. Draw a circle of radius 3.2 cm. Draw a chord AB of this circle such that AB = 5 cm. Shade the minor segment of the circle.
Answer: Complete the following construction steps: First, draw a circle with centre O and radius equal to 3.2 cm. Select any point A on the circle. Using A as a centre point, draw an arc with radius 5 cm that intersects the circle at another point B. Connect A and B with a straight line to form the chord. Finally, shade the region between the chord AB and the minor arc - this is the minor segment.
In simple words: Draw a circle. Mark two points on it 5 cm apart. The smaller piece cut off by the line joining them is the minor segment.

Exam Tip: Ensure the radius of the arc (5 cm) is greater than the distance from A to the centre, otherwise the arc won't intersect the circle a second time. Shade clearly to distinguish the minor segment from the rest.

Question 3. Find the length of the tangent drawn to a circle of radius 3 cm, from a point at a distance 5 cm from the centre.
Answer: Let PT be the tangent from external point P to a circle with centre C. We know that CT = 3 cm (radius) and CP = 5 cm (distance from centre to point). Since a radius is always perpendicular to the tangent at the point of contact, angle CTP = 90°. In the right triangle CPT, apply the Pythagorean theorem: CP² = PT² + CT². Substituting the values: (5)² = PT² + (3)², which gives 25 = PT² + 9. Therefore, PT² = 16, and PT = 4 cm. The length of the tangent is 4 cm.
In simple words: A radius meets a tangent at a right angle. Use Pythagoras' rule to find the tangent length.

Exam Tip: Always identify the right angle formed between the radius and tangent. This allows you to apply Pythagoras' theorem directly to find the missing side.

Question 4. In the adjoining figure, PT is a tangent to the circle with centre C. Given CP = 20 cm and PT = 16 cm, find the radius of the circle.
Answer: Since a radius is perpendicular to the tangent at the point of touch, we have CT ⊥ PT, which means angle CTP = 90°. This makes triangle CPT a right-angled triangle where CP is the hypotenuse. Using Pythagoras' theorem: CP² = PT² + CT². Rearranging, CT² = CP² - PT². Substitute the given values: CT² = (20)² - (16)² = 400 - 256 = 144. Taking the square root: CT = √144 = 12 cm. Therefore, the radius of the circle is 12 cm.
In simple words: The distance from the centre to an external point is the hypotenuse of a right triangle. Find the radius using Pythagoras' theorem.

Exam Tip: Always remember that CT is the hypotenuse when finding the radius - don't confuse which side is the hypotenuse in the right triangle.

Question 5. In each of the following figure, O is the centre of the circle. Find the size of each lettered angle:
(i) In the given figure, AB is the diameter and O is the centre. Given ∠CAB = 32° and ∠ABD = 50°. Since angles inscribed in a semicircle measure 90°, we have ∠C = 90°.
Answer: For angle x (which is ∠ABC): Using the angle sum property of triangles in triangle ABC: ∠C + ∠CAB + ∠ABC = 180°. Substitute: 90° + 32° + ∠x = 180°. This gives ∠x = 180° - 90° - 32° = 58°. For angle y (which is ∠BAD): In right-angled triangle ADB, we have ∠ABD + ∠ADB + ∠BAD = 180°. This becomes 50° + 90° + ∠y = 180°. Therefore, ∠y = 180° - 50° - 90° = 40°. Thus, x = 58° and y = 40°.
In simple words: Use the fact that angles in a semicircle are 90°. Then apply the angle sum rule for triangles to find the unknown angles.

Exam Tip: Always identify which angles are in a semicircle first - they must equal 90°. This is often the key to solving the problem quickly.

Question 5. (ii) In the figure, AC is the diameter with centre O. Given ∠DAC = 37° and AD is parallel to BC.
Answer: Since AD is parallel to BC, angles DAC and ACB are alternate angles, so ∠ACB = ∠DAC = 37°. Therefore, x = 37°. In triangle ABC, the angle inscribed in a semicircle gives ∠B = 90°. Using the angle sum property: ∠x + ∠y + ∠B = 180°. Substitute: 37° + ∠y + 90° = 180°. This gives ∠y = 180° - 127° = 53°. Thus, x = 37° and y = 53°.
In simple words: When two lines are parallel, alternate angles are equal. Then use the triangle angle sum to find the second angle.

Exam Tip: Recognise parallel lines in the figure immediately - they help you find equal angles without calculation.

Question 5. (iii) In the figure, AC is the diameter and O is the centre. BA = BC.
Answer: Since BA = BC, triangle ABC is isosceles. In an isosceles triangle, the base angles are equal, so ∠BAC = ∠BCA. Also, any angle inscribed in a semicircle equals 90°, so ∠ABC = 90°. Using the angle sum property: ∠BAC + ∠ABC + ∠BCA = 180°. Since ∠BAC = ∠BCA, let both equal x. Then: x + 90° + x = 180°. This gives 2x = 90°, so x = 45°.
In simple words: Equal sides make equal angles. The right angle at B plus two equal angles must sum to 180°.

Exam Tip: Spot isosceles triangles quickly by checking if two sides are marked equal - this immediately tells you two angles are also equal.

Question 5. (iv) In the figure, AC is the diameter with centre O. Given ∠ACD = 122°.
Answer: Angles ACB and ACD form a linear pair (they are on a straight line), so ∠ACB + ∠ACD = 180°. Therefore, ∠ACB = 180° - 122° = 58°. In triangle ABC, the angle in a semicircle gives ∠ABC = 90°. Using the angle sum property: ∠ABC + ∠BCA + ∠CAB = 180°. Substitute: 90° + 58° + x = 180°. This gives x = 180° - 148° = 32°.
In simple words: Angles on a straight line add to 180°. Then use the triangle sum to find the remaining angle.

Exam Tip: Always identify linear pairs - supplementary angles are often the starting point for solving these problems.

Question 5. (v) In the figure, AC is the diameter with centre O. OD is parallel to CB and ∠CAB = 40°.
Answer: In triangle ABC, angle B equals 90° (angle in a semicircle). Using the angle sum property: ∠BCA + ∠ABC + ∠BAC = 180°. This becomes ∠BCA + 90° + 40° = 180°, giving ∠BCA = 50°. Therefore, x = 50°. Since OD is parallel to CB, angles AOD and BCA are corresponding angles, so ∠AOD = 50°. Angles AOD and DOC form a linear pair on the straight line AC, so ∠AOD + ∠DOC = 180°. This gives 50° + y = 180°, so y = 130°. Thus, x = 50° and y = 130°.
In simple words: Find one angle using the triangle rule. Then use parallel lines to find another angle. Finally, use the straight line rule.

Exam Tip: Work through parallel line properties before linear pair properties - it often requires fewer steps.

Question 5. (vi) In the figure, AC is the diameter with centre O. BA = BC = CD.
Answer: In triangle ABC, angle B equals 90° (angle in a semicircle). Since BA = BC, the triangle is isosceles, so ∠BAC = ∠BCA = x. Using angle sum: x + 90° + x = 180°, giving 2x = 90°, so x = 45°. In triangle BCD, since BC = CD, it is isosceles, so ∠CBD = ∠CDB = y. By the exterior angle theorem, the exterior angle ∠ACB equals the sum of the two non-adjacent interior angles: ∠ACB = ∠CBD + ∠CDB. This gives x = y + y, or 45° = 2y, so y = 22.5°.
In simple words: Find the angle in the first isosceles triangle. Then use the exterior angle rule to find angles in the second triangle.

Exam Tip: Use the exterior angle theorem for efficiency - it often avoids the need to find all three angles in a triangle.

Question 5. (vii) In the figure, AB is the diameter of the circle with centre O. ST is the tangent at point B and ∠ASB = 65°.
Answer: Since ST is tangent to the circle at B and OB is a radius, they are perpendicular, giving ∠ABS = 90°. In triangle ABS, using the angle sum property: ∠BAC + ∠ASB + ∠ABS = 180°. Substitute: x + 65° + 90° = 180°. This gives x + 155° = 180°, so x = 25°. Therefore, x = 25°.
In simple words: A tangent meets the radius at a right angle. Use this to find the third angle in the triangle.

Exam Tip: Always mark the 90° angle formed between a tangent and radius - it's crucial for solving the problem.

Question 5. (viii) In the figure, AB is the diameter of the circle with centre O. ST is the tangent at point B and AB = BS.
Answer: Since ST is tangent at B and OB is the radius, angle OBS = 90°. In triangle ABS, using the angle sum property: ∠BAS + ∠BSA + ∠ABS = 180°. This becomes x + y + 90° = 180°, giving x + y = 90°. Since AB = BS, triangle ABS is isosceles, so x = y. Therefore, 2x = 90°, which gives x = 45°. Similarly, y = 45°, or y = 22.5° (if calculating as x/2 from a different configuration). Thus, x = y = 45°.
In simple words: Equal sides in a triangle mean equal angles. The right angle plus two equal angles sum to 180°.

Exam Tip: Isosceles triangles formed with tangents are common - always check for equal sides first.

Question 5. (ix) In the figure, RS is the diameter of the circle with centre O. SR is extended to Q, QT is the tangent at P, and ∠Q = 36°.
Answer: Since QT is tangent and OP is the radius, angle OPQ = 90°. In triangle OPQ, using the angle sum property: ∠OQP + ∠POQ + ∠OPQ = 180°. Substitute: 36° + ∠POQ + 90° = 180°, giving ∠POQ = 54°. Therefore, x = 54°. In triangle OPS, since OP = OS (both radii), the triangle is isosceles, so ∠OPS = ∠OSP = y. By the exterior angle theorem, ∠POQ = ∠OPS + ∠OSP, giving 54° = 2y. Therefore, y = 27°. Thus, x = 54° and y = 27°, or x = 2y = 54°.
In simple words: Find the angle at the centre using the triangle sum. Then use the isosceles triangle property with the exterior angle rule.

Exam Tip: The exterior angle equals the sum of the two remote interior angles - use this to avoid calculating all angles separately.

Question 6. In each of the following figures, O is the centre of the circle. Find the values of x and y.
(i) Given: O is the centre, AB = 15 cm, BC = 8 cm
Answer: Since AB is the diameter and C lies on the circle, angle ACB = 90° (angle in a semicircle). Using Pythagoras' theorem on triangle ABC: AC² = AB² + BC². Substitute: AC² = (15)² + (8)² = 225 + 64 = 289. Therefore, AC = √289 = 17 cm. Since AC is the diameter, x = 17 cm. The radius y equals half the diameter: y = AC/2 = 17/2 = 8.5 cm. Thus, x = 17 cm and y = 8.5 cm.
In simple words: The angle at C is 90° because it's in a semicircle. Find the diameter using Pythagoras. The radius is half the diameter.

Exam Tip: Always use the angle in a semicircle property - it's 90°. Then apply Pythagoras' theorem to find the hypotenuse (diameter).

Question 6. (ii) Given: O is the centre, PT and PS are tangents from P, OS = OT = 5 cm, PT = 12 cm
Answer: Since PT and PS are tangents from external point P, and OS and OT are radii, the angles OSP and OTP are both 90° (radius perpendicular to tangent). Also, PT = PS = 12 cm (tangents from an external point are equal). In right triangle OTP, apply Pythagoras' theorem: OP² = OT² + PT² = (5)² + (12)² = 25 + 144 = 169. Therefore, OP = √169 = 13 cm. So, x = 13 cm. Since tangents from P are equal, y = PT = 12 cm. Thus, x = 13 cm and y = 12 cm.
In simple words: Tangents from an external point are equal. The radius meets the tangent at a right angle. Use Pythagoras to find the distance from P to the centre.

Exam Tip: Tangent properties (equal lengths, perpendicular to radius) are key. Apply them before reaching for Pythagoras' theorem.

Question 6. (iii) Given: O is the centre, OT₁ = 24 cm (radius), PT₁ = 18 cm (tangent)
Answer: Since OT₁ is the radius and PT₁ is the tangent, angle OT₁P = 90°. In right triangle OPT₁, apply Pythagoras' theorem: OP² = OT₁² + PT₁² = (24)² + (18)² = 576 + 324 = 900. Therefore, OP = √900 = 30 cm. So, x = 30 cm. Since PT₁ and PT₂ are tangents from the same external point P to the circle, they are equal: PT₁ = PT₂ = 18 cm. Therefore, y = 18 cm. Thus, x = 30 cm and y = 18 cm.
In simple words: Use Pythagoras with the right angle formed between the radius and tangent. Tangents from one point are always equal.

Exam Tip: Mark the 90° angle at the tangent point clearly. This makes applying Pythagoras' theorem straightforward.