Circles, Centres & Collinearity — Class 9 Mathematics Chapter 1

🖼 Image PendingA wide cinematic vista: ripples spreading outward from a single dropped pebble at the centre of a still pond, forming concentric circles of light, with three distant lamps along the far shore appearing to lie on a single line on the horizon

AI Generation Prompt

Ultra-wide cinematic banner (16:5 ratio). A still pond at twilight. A single pebble has just struck the centre of the water and concentric circles of golden ripples spread outward in perfect symmetry. In the far distance, on the opposite shore, three small lamps glow on the horizon — appearing to lie almost exactly on a single straight line. The image conveys two ideas at once: the *circle* as the locus of points equidistant from a centre, and the *line* as a special locus of three points that happen to fall together. Painterly cinematic illustration. Dark background. No text, no labels.

✦आर्यभटीय — २.७ (वृत्तं समकेन्द्रम्)

From the Āryabhaṭīya — On Circles

वृत्तं समकेन्द्रं समदूरम् च परिधितः।

केन्द्रात् सर्वे बिन्दवः समदूरास्थिताः ज्ञेयाः॥

(vṛttaṃ samakendraṃ samadūram ca paridhitaḥ / kendrāt sarve bindavaḥ samadūrāsthitāḥ jñeyāḥ)

'गोल आकृति का एक केंद्र होता है — और हर बिन्दु जो उस आकृति पर है, केंद्र से बराबर दूरी पर है।'

'A circle has a single centre, and every point on it is at the same distance from that centre.'

Āryabhaṭa (499 CE) made the locus definition of a circle explicit: the circle is the set of points equidistant from a given centre. This page turns that classical definition into a coordinate equation.

A circle is a locus

A locus (Latin for place) is the set of all points satisfying a given condition.

Definition. The circle of centre $(h, k)$ and radius $r$ is the set of all points $(x, y)$ in the plane whose distance from $(h, k)$ equals exactly $r$ .

Using the distance formula, this means:

$\sqrt{(x - h)^2 + (y - k)^2} = r$

Squaring both sides (both sides are non-negative, so squaring is reversible):

$\boxed{(x - h)^2 + (y - k)^2 = r^2}$

This is the equation of a circle in standard form. Memorise it.

Special case — circle centred at the origin. When $h = k = 0$ :

$x^2 + y^2 = r^2$

Three things you can ask given a circle and a point $P(x_0, y_0)$ :

Is $P$ on the circle? Compute $(x_0 - h)^2 + (y_0 - k)^2$ . If it equals $r^2$ , yes.
Is $P$ inside the circle? Same expression — if it is less than $r^2$ , $P$ is inside.
Is $P$ outside? If the expression is greater than $r^2$ , $P$ is outside.

This classification is purely a comparison — no square roots needed.

Loading simulator…

NCERT Q12 — A circle through three points

Worked Example 1 (NCERT Q12 i)Three points on a circle

NCERT

Show that the points $A(1, -8)$ , $B(-4, 7)$ , $C(-7, -4)$ all lie on a circle $K$ whose centre is the origin $O(0, 0)$ . What is the radius of circle $K$ ?

Solution

Strategy. A point $(x, y)$ lies on the circle of centre $O(0, 0)$ and radius $r$ exactly when $\sqrt{x^2 + y^2} = r$ — i.e., when the distance from the origin equals $r$ . So all three points lie on a single circle centred at the origin if and only if they are all the same distance from the origin.

Step 1 — Distance from origin to each point.

$OA = \sqrt{1^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$
$OB = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$
$OC = \sqrt{(-7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}$

Step 2 — Compare.

All three distances equal $\sqrt{65}$ . So all three points lie on the circle of radius $\sqrt{65}$ centred at $O$ .

Answer: Radius of circle $K$ is $\sqrt{65}$ units (≈ 8.06 units). The equation of the circle is $x^2 + y^2 = 65$ .

Aesthetic note. Since $1 + 64 = 16 + 49 = 49 + 16 = 65$ , the three points form a triple where each pair $(x, y)$ has $x^2 + y^2 = 65$ . There is something pleasing about a single number ( $65$ ) hiding three different ways to be split as a sum of squares.

Worked Example 2 (NCERT Q12 ii)Inside, on, or outside?

NCERT

Given the points $D(-5, 6)$ and $E(0, 9)$ , check whether each lies inside, on, or outside the circle $K$ from the previous problem.

Solution

Recall. Circle $K$ has centre $O(0, 0)$ and radius $\sqrt{65}$ — i.e., $r^2 = 65$ . To classify a point $(x, y)$ :

$x^2 + y^2 = 65$ → on the circle
$x^2 + y^2 < 65$ → inside
$x^2 + y^2 > 65$ → outside

Step 1 — Test $D(-5, 6)$ .

$D_x^2 + D_y^2 = (-5)^2 + 6^2 = 25 + 36 = 61$

$61 < 65$ ⇒ $D$ lies inside the circle.

Step 2 — Test $E(0, 9)$ .

$E_x^2 + E_y^2 = 0^2 + 9^2 = 0 + 81 = 81$

$81 > 65$ ⇒ $E$ lies outside the circle.

Answer: $D$ is inside circle $K$ ; $E$ is outside.

Note: We did not need to take any square roots. Comparing $x^2 + y^2$ to $r^2$ directly is cleaner and avoids irrational numbers.

NCERT Q15 — Computer-graphics circles

Worked Example 3 (NCERT Q15)Two icons on a screen

NCERT

A computer screen is 800 pixels wide and 600 pixels high, with the origin at the bottom-left corner. A circular icon of radius 80 pixels is drawn with centre $A(100, 150)$ . Another circular icon of radius 100 pixels is drawn with centre $B(250, 230)$ .

(i) Determine whether any part of either circle lies outside the screen.

(ii) Determine whether the two circles intersect each other.

Solution

Part (i) — Off-screen check.

A circle of centre $(h, k)$ and radius $r$ lies entirely on screen (with screen $0 \le x \le 800$ , $0 \le y \le 600$ ) when:

$h - r \ge 0$ (left edge)
$h + r \le 800$ (right edge)
$k - r \ge 0$ (bottom edge)
$k + r \le 600$ (top edge)

Icon A ( $h=100, k=150, r=80$ ):

Left edge: $100 - 80 = 20 \ge 0$ ✓
Right edge: $100 + 80 = 180 \le 800$ ✓
Bottom edge: $150 - 80 = 70 \ge 0$ ✓
Top edge: $150 + 80 = 230 \le 600$ ✓

Icon A lies entirely on screen. ✓

Icon B ( $h=250, k=230, r=100$ ):

Left edge: $250 - 100 = 150 \ge 0$ ✓
Right edge: $250 + 100 = 350 \le 800$ ✓
Bottom edge: $230 - 100 = 130 \ge 0$ ✓
Top edge: $230 + 100 = 330 \le 600$ ✓

Icon B also lies entirely on screen. ✓

Part (ii) — Do the circles intersect?

Two circles intersect if and only if the distance between their centres is less than the sum of the radii AND greater than the absolute difference of the radii.

Sum of radii: $80 + 100 = 180$
Difference of radii: $|100 - 80| = 20$
Distance between centres: $\sqrt{(250 - 100)^2 + (230 - 150)^2} = \sqrt{150^2 + 80^2} = \sqrt{22500 + 6400} = \sqrt{28900}$

Simplify: $\sqrt{28900} = \sqrt{100 \cdot 289} = 10 \cdot 17 = 170$ .

Compare. $20 < 170 < 180$ — distance is between the difference and the sum, so the two circles intersect in two points.

Answer:

(i) Both icons lie entirely on the screen — no off-screen parts.
(ii) The two circles intersect (their boundaries cross at two points).

Collinearity — when three points lie on a line

Three points $A$ , $B$ , $C$ are collinear when they all lie on a single straight line.

The distance test. Three points are collinear if and only if the largest of the three pair-distances equals the sum of the other two.

Why? When the three points are not collinear, they form a triangle. The triangle inequality says that the largest side is less than the sum of the other two — there is room for a real triangle. When the three points are collinear, the triangle has degenerated to a line: the longest 'side' (between the two extreme points) is exactly the sum of the other two segments.

So the test is:

$\text{collinear} \iff \max(AB, BC, AC) = \text{sum of the other two}$

This is one of several collinearity tests; you'll meet a slope-based test in Class 10. The distance-based test has the advantage of using only what you already know.

Worked Example 4 (NCERT Q6)Are three points collinear?

NCERT

Are the points $M(-3, -4)$ , $A(0, 0)$ , $G(6, 8)$ on the same straight line? Suggest a method that does not require plotting and joining the points.

Solution

Method — distance test.

Compute all three pair-distances and check whether the largest equals the sum of the other two.

Step 1 — Compute the three distances.

$MA = \sqrt{(0 - (-3))^2 + (0 - (-4))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
$AG = \sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
$MG = \sqrt{(6 - (-3))^2 + (8 - (-4))^2} = \sqrt{81 + 144} = \sqrt{225} = 15$

Step 2 — Apply the test.

Largest distance: $MG = 15$ . Sum of the other two: $MA + AG = 5 + 10 = 15$ .

$15 = 15$ ✓

Conclusion. The three points are collinear — they lie on a single straight line, with $A$ between $M$ and $G$ .

Answer: Yes, $M$ , $A$ , $G$ are collinear. The method (distance test) uses only the distance formula and works without any plot.

Worked Example 5 (NCERT Q7)Apply the same method

NCERT

Use the distance test to check whether $R(-5, -1)$ , $B(-2, -5)$ , $C(4, -12)$ are on the same straight line.

Solution

Step 1 — Compute the three distances.

$RB = \sqrt{(-2 - (-5))^2 + (-5 - (-1))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
$BC = \sqrt{(4 - (-2))^2 + (-12 - (-5))^2} = \sqrt{36 + 49} = \sqrt{85}$
$RC = \sqrt{(4 - (-5))^2 + (-12 - (-1))^2} = \sqrt{81 + 121} = \sqrt{202}$

Step 2 — Apply the test.

Largest distance: $RC = \sqrt{202} \approx 14.21$ . Sum of the other two: $RB + BC = 5 + \sqrt{85} \approx 5 + 9.22 = 14.22$ .

These are almost equal but not exactly equal. Let me check more carefully: is $\sqrt{202}$ exactly equal to $5 + \sqrt{85}$ ?

Squaring: $(5 + \sqrt{85})^2 = 25 + 10\sqrt{85} + 85 = 110 + 10\sqrt{85}$ . And $202 \ne 110 + 10\sqrt{85}$ (since $10\sqrt{85} \approx 92.2$ , so $110 + 92.2 = 202.2 \ne 202$ ).

Conclusion. The three points are not exactly collinear — they are very nearly on a line, but not quite. They form a very thin (nearly degenerate) triangle.

Answer: No, $R$ , $B$ , $C$ are not on the same straight line. The pair-distances almost satisfy the collinearity test, but not exactly — a useful reminder to compute distances exactly (with $\sqrt{}$ symbols), not just decimal approximations, before declaring collinearity.

Practice Yourself — Circles & Collinearity

A. Equation of a circle. Write the equation of:

The circle centred at the origin with radius 7.
The circle centred at $(3, -4)$ with radius 5.
The unit circle (centre origin, radius 1).
A circle that passes through $(3, 4)$ and is centred at the origin.

B. Inside, on, or outside. Each of these refers to the circle $x^2 + y^2 = 100$ . Classify each point.

$(6, 8)$
$(7, 7)$
$(0, 11)$
$(-6, -8)$

C. Collinearity. Use the distance test to decide whether each triple is collinear.

$(0, 0)$ , $(3, 4)$ , $(6, 8)$
$(1, 2)$ , $(3, 4)$ , $(5, 5)$
$(-2, -3)$ , $(0, 0)$ , $(2, 3)$
$(0, 0)$ , $(1, 2)$ , $(2, 5)$

Answers: 1. $x^2 + y^2 = 49$ . 2. $(x - 3)^2 + (y + 4)^2 = 25$ . 3. $x^2 + y^2 = 1$ . 4. $x^2 + y^2 = 25$ . 5. on (36 + 64 = 100). 6. inside (49 + 49 = 98 < 100). 7. outside (0 + 121 > 100). 8. on (36 + 64 = 100). 9. Collinear ( $d_1 = 5$ , $d_2 = 5$ , $d_3 = 10$ ✓). 10. Not collinear. 11. Collinear ( $d_1 = \sqrt{13}$ , $d_2 = \sqrt{13}$ , $d_3 = \sqrt{52} = 2\sqrt{13}$ ✓). 12. Not collinear.

Ready to Go Beyond

The locus idea generalises beautifully. A circle is the locus of points at fixed distance from one centre. What about more centres?

Quick Check

Q1.What is the equation of the circle centred at the origin with radius 6?

AI Generation Prompt

✦आर्यभटीय — २.७ (वृत्तं समकेन्द्रम्)

From the Āryabhaṭīya — On Circles

वृत्तं समकेन्द्रं समदूरम् च परिधितः।

केन्द्रात् सर्वे बिन्दवः समदूरास्थिताः ज्ञेयाः॥

(vṛttaṃ samakendraṃ samadūram ca paridhitaḥ / kendrāt sarve bindavaḥ samadūrāsthitāḥ jñeyāḥ)

'A circle has a single centre, and every point on it is at the same distance from that centre.'

A circle is a locus

A locus (Latin for place) is the set of all points satisfying a given condition.

Definition. The circle of centre $(h, k)$ and radius $r$ is the set of all points $(x, y)$ in the plane whose distance from $(h, k)$ equals exactly $r$ .

Using the distance formula, this means:

$\sqrt{(x - h)^2 + (y - k)^2} = r$

Squaring both sides (both sides are non-negative, so squaring is reversible):

$\boxed{(x - h)^2 + (y - k)^2 = r^2}$

This is the equation of a circle in standard form. Memorise it.

Special case — circle centred at the origin. When $h = k = 0$ :

$x^2 + y^2 = r^2$

Three things you can ask given a circle and a point $P(x_0, y_0)$ :

Is $P$ on the circle? Compute $(x_0 - h)^2 + (y_0 - k)^2$ . If it equals $r^2$ , yes.
Is $P$ inside the circle? Same expression — if it is less than $r^2$ , $P$ is inside.
Is $P$ outside? If the expression is greater than $r^2$ , $P$ is outside.

This classification is purely a comparison — no square roots needed.

Loading simulator…

NCERT Q12 — A circle through three points

Worked Example 1 (NCERT Q12 i)Three points on a circle

NCERT

Show that the points $A(1, -8)$ , $B(-4, 7)$ , $C(-7, -4)$ all lie on a circle $K$ whose centre is the origin $O(0, 0)$ . What is the radius of circle $K$ ?

Solution

Step 1 — Distance from origin to each point.

$OA = \sqrt{1^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$
$OB = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$
$OC = \sqrt{(-7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}$

Step 2 — Compare.

All three distances equal $\sqrt{65}$ . So all three points lie on the circle of radius $\sqrt{65}$ centred at $O$ .

Answer: Radius of circle $K$ is $\sqrt{65}$ units (≈ 8.06 units). The equation of the circle is $x^2 + y^2 = 65$ .

Worked Example 2 (NCERT Q12 ii)Inside, on, or outside?

NCERT

Given the points $D(-5, 6)$ and $E(0, 9)$ , check whether each lies inside, on, or outside the circle $K$ from the previous problem.

Solution

Recall. Circle $K$ has centre $O(0, 0)$ and radius $\sqrt{65}$ — i.e., $r^2 = 65$ . To classify a point $(x, y)$ :

$x^2 + y^2 = 65$ → on the circle
$x^2 + y^2 < 65$ → inside
$x^2 + y^2 > 65$ → outside

Step 1 — Test $D(-5, 6)$ .

$D_x^2 + D_y^2 = (-5)^2 + 6^2 = 25 + 36 = 61$

$61 < 65$ ⇒ $D$ lies inside the circle.

Step 2 — Test $E(0, 9)$ .

$E_x^2 + E_y^2 = 0^2 + 9^2 = 0 + 81 = 81$

$81 > 65$ ⇒ $E$ lies outside the circle.

Answer: $D$ is inside circle $K$ ; $E$ is outside.

Note: We did not need to take any square roots. Comparing $x^2 + y^2$ to $r^2$ directly is cleaner and avoids irrational numbers.

NCERT Q15 — Computer-graphics circles

Worked Example 3 (NCERT Q15)Two icons on a screen

NCERT

(i) Determine whether any part of either circle lies outside the screen.

(ii) Determine whether the two circles intersect each other.

Solution

Part (i) — Off-screen check.

A circle of centre $(h, k)$ and radius $r$ lies entirely on screen (with screen $0 \le x \le 800$ , $0 \le y \le 600$ ) when:

$h - r \ge 0$ (left edge)
$h + r \le 800$ (right edge)
$k - r \ge 0$ (bottom edge)
$k + r \le 600$ (top edge)

Icon A ( $h=100, k=150, r=80$ ):

Left edge: $100 - 80 = 20 \ge 0$ ✓
Right edge: $100 + 80 = 180 \le 800$ ✓
Bottom edge: $150 - 80 = 70 \ge 0$ ✓
Top edge: $150 + 80 = 230 \le 600$ ✓

Icon A lies entirely on screen. ✓

Icon B ( $h=250, k=230, r=100$ ):

Left edge: $250 - 100 = 150 \ge 0$ ✓
Right edge: $250 + 100 = 350 \le 800$ ✓
Bottom edge: $230 - 100 = 130 \ge 0$ ✓
Top edge: $230 + 100 = 330 \le 600$ ✓

Icon B also lies entirely on screen. ✓

Part (ii) — Do the circles intersect?

Two circles intersect if and only if the distance between their centres is less than the sum of the radii AND greater than the absolute difference of the radii.

Sum of radii: $80 + 100 = 180$
Difference of radii: $|100 - 80| = 20$
Distance between centres: $\sqrt{(250 - 100)^2 + (230 - 150)^2} = \sqrt{150^2 + 80^2} = \sqrt{22500 + 6400} = \sqrt{28900}$

Simplify: $\sqrt{28900} = \sqrt{100 \cdot 289} = 10 \cdot 17 = 170$ .

Compare. $20 < 170 < 180$ — distance is between the difference and the sum, so the two circles intersect in two points.

Answer:

(i) Both icons lie entirely on the screen — no off-screen parts.
(ii) The two circles intersect (their boundaries cross at two points).

Collinearity — when three points lie on a line

Three points $A$ , $B$ , $C$ are collinear when they all lie on a single straight line.

The distance test. Three points are collinear if and only if the largest of the three pair-distances equals the sum of the other two.

So the test is:

$\text{collinear} \iff \max(AB, BC, AC) = \text{sum of the other two}$

This is one of several collinearity tests; you'll meet a slope-based test in Class 10. The distance-based test has the advantage of using only what you already know.

Worked Example 4 (NCERT Q6)Are three points collinear?

NCERT

Are the points $M(-3, -4)$ , $A(0, 0)$ , $G(6, 8)$ on the same straight line? Suggest a method that does not require plotting and joining the points.

Solution

Method — distance test.

Compute all three pair-distances and check whether the largest equals the sum of the other two.

Step 1 — Compute the three distances.

$MA = \sqrt{(0 - (-3))^2 + (0 - (-4))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
$AG = \sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
$MG = \sqrt{(6 - (-3))^2 + (8 - (-4))^2} = \sqrt{81 + 144} = \sqrt{225} = 15$

Step 2 — Apply the test.

Largest distance: $MG = 15$ . Sum of the other two: $MA + AG = 5 + 10 = 15$ .

$15 = 15$ ✓

Conclusion. The three points are collinear — they lie on a single straight line, with $A$ between $M$ and $G$ .

Answer: Yes, $M$ , $A$ , $G$ are collinear. The method (distance test) uses only the distance formula and works without any plot.

Worked Example 5 (NCERT Q7)Apply the same method

NCERT

Use the distance test to check whether $R(-5, -1)$ , $B(-2, -5)$ , $C(4, -12)$ are on the same straight line.

Solution

Step 1 — Compute the three distances.

$RB = \sqrt{(-2 - (-5))^2 + (-5 - (-1))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
$BC = \sqrt{(4 - (-2))^2 + (-12 - (-5))^2} = \sqrt{36 + 49} = \sqrt{85}$
$RC = \sqrt{(4 - (-5))^2 + (-12 - (-1))^2} = \sqrt{81 + 121} = \sqrt{202}$

Step 2 — Apply the test.

Largest distance: $RC = \sqrt{202} \approx 14.21$ . Sum of the other two: $RB + BC = 5 + \sqrt{85} \approx 5 + 9.22 = 14.22$ .

These are almost equal but not exactly equal. Let me check more carefully: is $\sqrt{202}$ exactly equal to $5 + \sqrt{85}$ ?

Squaring: $(5 + \sqrt{85})^2 = 25 + 10\sqrt{85} + 85 = 110 + 10\sqrt{85}$ . And $202 \ne 110 + 10\sqrt{85}$ (since $10\sqrt{85} \approx 92.2$ , so $110 + 92.2 = 202.2 \ne 202$ ).

Conclusion. The three points are not exactly collinear — they are very nearly on a line, but not quite. They form a very thin (nearly degenerate) triangle.

Ready to Go Beyond

The locus idea generalises beautifully. A circle is the locus of points at fixed distance from one centre. What about more centres?

Quick Check

Q1.What is the equation of the circle centred at the origin with radius 6?