NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.1#

Theorem 1: There is a unique circle passing through three non-collinear points.

Its centre, the circumcentre, is where the perpendicular bisectors of the sides meet, because every point on the perpendicular bisector of $AB$ is equally far from $A$ and $B$.

To draw the circumcircle of a triangle: draw the perpendicular bisectors of any two sides; they meet at the circumcentre $O$. With centre $O$ and radius $OA$, draw the circle. It passes through all three vertices.

1. Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside 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. Draw the perpendicular bisectors of $AB$ and $AC$. Let them meet at $O$.
  4. With centre $O$ and radius $OA$, draw the circle. It passes through $A$, $B$ and $C$.
Inside or outside?

The third angle is:

$$\angle C = 180° - 70° - 60°$$

$$\angle C = 50°$$

All three angles (70°, 60°, 50°) are acute. In an acute-angled triangle the perpendicular bisectors meet inside the triangle.

∴ The circumcentre $O$ is inside the triangle, as the figure shows.

2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle? #

Construction
  1. Draw $AB = 5$ cm.
  2. At $A$, draw a ray making 100° with $AB$, and mark $C$ on it with $AC = 4$ cm. Join $BC$.
  3. Draw the perpendicular bisectors of $AB$ and $AC$. Let them meet at $O$.
  4. With centre $O$ and radius $OA$, draw the circle through $A$, $B$ and $C$.
Inside or outside?

$\angle A = 100°$ is obtuse. In an obtuse-angled triangle the perpendicular bisectors meet outside the triangle, beyond the longest side $BC$ (the side opposite the obtuse angle).

∴ The circumcentre $O$ is outside the triangle, as the figure shows.

3. 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. #

Construction
  1. Draw $AB = 6$ cm.
  2. With centre $A$ and radius 7 cm, draw an arc; with centre $B$ and radius 7 cm, draw another arc cutting it at $C$. Join $AC$ and $BC$.
  3. Draw the perpendicular bisectors of $AB$ and $BC$. Let them meet at $O$.
  4. With centre $O$ and radius $OA$, draw the circumcircle.
Measurement

Measuring, we find

$$OA = OB = OC \approx 3.9 \text{ cm}$$

All three are equal, because $O$ is the centre and $OA$, $OB$, $OC$ are radii of the circumcircle.

(Calculating exactly gives $3.87$ cm, so a measurement of about 3.9 cm is correct.)

4. What is the least possible radius of a circle through two points A and B? #

Solution

The centre $O$ of any circle through $A$ and $B$ is equally far from $A$ and $B$, so $O$ lies on the perpendicular bisector of $AB$. Let $M$ be the midpoint of $AB$.

If $O$ is not at $M$, then $\triangle OMA$ is right-angled at $M$ and $OA$ is its hypotenuse, so:

$$OA > AM$$

If $O$ is at $M$, then:

$$OA = AM$$

So the radius $OA$ is smallest when the centre is the midpoint $M$:

$$\text{Least radius} = AM = \dfrac{AB}{2}$$

∴ The least possible radius is $\dfrac{AB}{2}$ — the circle that has $AB$ as a diameter.

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