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NCERT Solutions for Class 9th Maths Chapter 5 Exercise Set 5.1 — How Many Circles?

Book page 98 Updated on2026-09-08

Q1.
Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
Answer

The centre lies inside the triangle.

Construction:

  1. Draw AB = 5 cm.
  2. At A draw a ray making 70° with AB; at B draw a ray making 60° with BA. They meet at C.
  3. Construct the perpendicular bisectors of AB and of BC. They meet at O.
  4. With centre O and radius OA, draw the circle. It passes through B and C as well.
∠C = 180° − 70° − 60° = 50°
Angles are 70°, 60°, 50° — all acute, so ΔABC is an acute-angled triangle.
For an acute-angled triangle the circumcentre lies inside the triangle.
Check it yourself: measure OA, OB, OC — all three should come out about 3.3 cm. (Exactly, R = AB / (2 sin C) = 5 / (2 sin 50°) ≈ 3.26 cm.)
Why it happens: the circumcentre O sees side AB under the angle ∠AOB = 2∠C. O falls inside the triangle exactly when all three of these central angles 2∠A, 2∠B, 2∠C are less than 180° — that is, when every angle of the triangle is less than 90°.
Q2.
Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
Answer

The centre lies outside the triangle — on the far side of BC from A.

Construction: draw AB = 5 cm; at A draw a ray making 100° with AB and cut off AC = 4 cm on it; join BC. Then draw the perpendicular bisectors of AB and AC; they meet at O, outside the triangle. Draw the circle with centre O and radius OA.

∠A = 100° > 90° ⇒ ΔABC is an obtuse-angled triangle.
For an obtuse-angled triangle the circumcentre lies outside the triangle,
beyond the side opposite the obtuse angle — here beyond BC.
Check it yourself: BC ≈ 6.9 cm and OA = OB = OC ≈ 3.5 cm.
(BC² = 5² + 4² − 2·5·4·cos 100° ≈ 47.9, so BC ≈ 6.92 cm; R = BC / (2 sin 100°) ≈ 3.52 cm.)
Why it happens: the central angle standing on BC is 2∠A = 200°, which is a reflex angle. So A lies on the minor arc BC, and the centre is pushed across BC to the other side, out of the triangle.
Q3.
Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.
Answer

OA = OB = OC ≈ 3.9 cm. All three must be equal — they are radii of the same circle.

Construction: draw AB = 6 cm; with centre A and radius 7 cm, and centre B and radius 7 cm, draw arcs cutting at C. Construct the perpendicular bisectors of AB and BC; they meet at O. Draw the circle with centre O, radius OA.

The triangle is isosceles (CA = CB = 7 cm), so it is symmetric about the perpendicular bisector of AB.
Height from C to AB = √(7² − 3²) = √40 ≈ 6.32 cm
Area = ½ × 6 × 6.32 ≈ 18.97 cm²
R = (abc) / (4 × area) = (6 × 7 × 7) / (4 × 18.97) = 294 / 75.9 ≈ 3.87 cm
Tip: the angles here are about 64.6°, 64.6° and 50.8° — all acute — so O falls inside the triangle, on the perpendicular bisector of AB, about 2.45 cm above AB.
Why OA = OB = OC: that is what "circumcentre" means. O is on the perpendicular bisector of AB (so OA = OB) and on the perpendicular bisector of BC (so OB = OC). The two facts together force all three to be equal, which is exactly why a single circle can pass through all three vertices.
Q4.
What is the least possible radius of a circle through two points A and B?
Answer

The least possible radius is ½ AB — half the distance between the two points.

AB is a chord of any such circle.
No chord can exceed the diameter, so AB ≤ 2r, i.e. r ≥ AB/2.
The value r = AB/2 is actually reached: take the midpoint M of AB as centre.
Then MA = MB = AB/2, and the circle with centre M, radius AB/2, passes through both.
Least radius = AB/2, and for that circle AB is a diameter.
Why it happens: the centre must sit on the perpendicular bisector of AB, and its distance to A is √(h² + (AB/2)²), where h is how far it has moved from the midpoint. This is smallest when h = 0. Any move away from the midpoint only lengthens the radius.
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