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Mathematics — Class 12 — L15: Circles

NIOS Code 311 · Module 4 · Co-ordinate Geometry

Notes extracted from NIOS Mathematics Course (311), Lesson 15 — Circles (ilovepdf_merged (6).pdf). Content covers sections 15.1–15.x.
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Overview — Circles (L15)

A circle is the set of points at fixed distance r (radius) from a fixed point (centre). This lesson fixes standard equations, general form, and basic problems (centre–radius, diameter form, condition for a line to meet a circle).

(x − h)² + (y − k)² = r²
Centre (h,k) · radius r > 0
r (h,k) (x−h)² + (y−k)² = r²
Standard circle centre (h,k) radius r

15.1 Standard equation

Circle with centre origin and radius r: x² + y² = r². General centre (h,k): expand to x² + y² + Dx + Ey + F = 0 under conditions so that radius is real.

x² + y² + Dx + Ey + F = 0
Centre (−D/2, −E/2) · r = ½√(D²+E²−4F) if ≥ 0
x y O I (+,+) II (−,+) III (−,−) IV (+,−)
Coordinate axes and quadrants

15.2 Diameter form

If AB is a diameter with A(x₁,y₁), B(x₂,y₂), any point P on the circle (except A,B endpoints as right-angle cases) satisfies the angle in a semicircle property. Equation:

(x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0
Diameter endpoints (x₁,y₁), (x₂,y₂)

15.3 Position of a point / line

  • Point inside, on, or outside the circle is tested by comparing distance from centre with r.
  • For line L = 0 and circle, use perpendicular distance from centre vs r: equal → tangent; less → secant; greater → no real intersection.
Common mistake: Forgetting the condition D²+E²−4F ≥ 0 when claiming a real circle from the general equation.

MCQ Quiz — L15 Circles

0 / 10 correct

Flashcards — L15

1 / 12

Golden Rules — L15 Circles

Most exam-important points from this chapter:

Know the definitions of L15

Start every answer with the key definition or standard form from Circles.

Write the formula first

Copy the formula, then substitute values; never jump to the number alone.

Watch conditions

Check domain, quadrant, non-zero denominators, and applicability of the theorem.

Master result 1

Memorise and apply: Distance between two points with given coordinates.

Master result 2

Memorise and apply: constant distance is called the radius of the circle.

(x−h)²+(y−k)²=r²
x²+y²+2gx+2fy+c=0
Centre (−g,−f), r=√(g²+f²−c)
x²+y²=r²
Tangent condition
Chord with midpoint

1. Formulas & Definitions

Full Ch 15 — Circles study guide: definitions, derivations, why each formula works, history, deep understanding, pencil diagrams, step-by-step use, and 4-level solved examples.

(x − h)² + (y − k)² = r²

Definition: Standard equation of a circle: centre (h,k), radius r > 0.

Derivation

Definition: set of points at fixed distance r from centre → distance formula squared.

Variables

Centre C(h,k) · radius r

Why it works

Every point on the circle is exactly r from C; squaring removes the root.

Historical context

Apollonius studied circles; Cartesian form is modern school standard.

Deep understanding

Expanding gives general form with equal x² and y² coefficients and no xy term.

2. Diagrams & Visuals

(x − h)² + (y − k)² = r² Centre (h,k) Radius r constant (x−h)²+(y−k)²=r²

Pencil sketch · labelled · step-by-step breakdown below

  1. Identify centre and radius
  2. Write (x−h)²+(y−k)²=r²
  3. Expand if general form needed
  4. Complete square to reverse

3. Solved Examples

Basic

Q: Centre (0,0), r=5.

Solution: x²+y²=25

Answer: x² + y² = 25

Intermediate

Q: Centre (2,−3), r=4.

Solution: (x−2)²+(y+3)²=16

Answer: (x−2)²+(y+3)²=16

Advanced

Q: x²+y²−4x+6y−3=0 → centre, r.

Solution: (x−2)²+(y+3)²=16 → C(2,−3), r=4

Answer: C(2,−3), r=4

Exam

Q: Does (3,4) lie on x²+y²=25?

Solution: 9+16=25

Answer: Yes, on the circle

Centre (−g, −f), r = √(g² + f² − c)

Definition: For x²+y²+2gx+2fy+c=0 (with g²+f²−c > 0).

Derivation

Complete the square: (x+g)²+(y+f)² = g²+f²−c.

Variables

g, f, c = coefficients · r real iff g²+f²−c ≥ 0

Why it works

Completing square shifts origin to the true centre.

Historical context

General second-degree circle equation standardised in 18th–19th c. algebra texts.

Deep understanding

If g²+f²−c=0 → point circle; if negative → no real circle.

2. Diagrams & Visuals

Centre (−g, −f), r = √(g² + f² − c) Complete square on x,y Centre (−g,−f) r² = g²+f²−c

Pencil sketch · labelled · step-by-step breakdown below

  1. Read g, f, c from equation
  2. Centre = (−g,−f)
  3. r = √(g²+f²−c)
  4. State if real

3. Solved Examples

Basic

Q: x²+y²+2x+0y−8=0.

Solution: g=1,f=0,c=−8 → C(−1,0), r=3

Answer: C(−1,0), r=3

Intermediate

Q: x²+y²−6x−8y=0.

Solution: g=−3,f=−4,c=0 → C(3,4), r=5

Answer: C(3,4), r=5

Advanced

Q: Find c so centre (1,−2), r=5: x²+y²−2x+4y+c=0.

Solution: 1+4−c=25 → c=−20

Answer: c = −20

Exam

Q: Condition for x²+y²+2gx+2fy+c=0 to represent a circle.

Solution: g²+f²−c > 0 (strict for positive r)

Answer: g²+f²−c > 0

T = 0 (tangent from point form)

Definition: For x²+y²=r², tangent at (x₁,y₁) on the circle is xx₁+yy₁=r².

Derivation

Differentiate implicitly or use radius ⊥ tangent at contact point.

Variables

(x₁,y₁) on circle · r = radius

Why it works

Radius to contact is normal; algebraic polar of the point becomes the tangent.

Historical context

Descartes and later algebraic geometry developed pole–polar for conics.

Deep understanding

For general circle replace by T=0 using S₁ notation from textbook.

2. Diagrams & Visuals

T = 0 (tangent from point form) Radius ⊥ tangent At contact (x₁,y₁) xx₁+yy₁=r²

Pencil sketch · labelled · step-by-step breakdown below

  1. Confirm point lies on circle
  2. Write xx₁+yy₁=r² (unit centre case)
  3. Generalise with centre if needed
  4. Check slope of radius · tangent = −1

3. Solved Examples

Basic

Q: Tangent to x²+y²=25 at (3,4).

Solution: 3x+4y=25

Answer: 3x + 4y = 25

Intermediate

Q: Tangent at (5,0) on x²+y²=25.

Solution: 5x=25 → x=5

Answer: x = 5

Advanced

Q: Condition that y=mx+c touch x²+y²=r².

Solution: c²=r²(1+m²)

Answer: c² = r²(1+m²)

Exam

Q: Find tangents from (0,0) wait origin inside unit circle?

Solution: Origin centre → no real tangent from centre

Answer: No real tangent from centre

5. Special Features & Extras

Complete study guide — exam tips, common mistakes, memory aids, and quick reference.

Exam Tips & Tricks

  • Always complete the square before reading centre/radius.
  • Equal coefficients of x² and y² and no xy ⇒ circle (or point/empty).
  • Tangent condition c² = r²(1+m²) for y=mx+c on x²+y²=r².

Common Student Mistakes

  • Writing centre (g,f) instead of (−g,−f)
  • Using r² = g²+f²+c with wrong sign
  • Forgetting point must lie on circle for tangent-at-point

Memory Aids & Mnemonics

Centre is opposite signs of half the linear terms.
r² = g² + f² − c (minus c).

Which Formula When?

  • Centre+radius known → standard form
  • General equation → complete square
  • Tangent at a point → T=0

Quick reference box

• (x−h)²+(y−k)²=r²

• C(−g,−f), r=√(g²+f²−c)

• xx₁+yy₁=r² on x²+y²=r²

PYQ — Previous Year Questions

English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.

L15 — Circles

4 question(s) · Sources: Board-style (from papers on circle centre), Board-style diameter form, Sample QP 2024

PYQ1. Find the equation of the circle whose centre is (3, 4) and radius is 5.

1 mark(s) · SA · Sample QP 2024 · Q3(ii) OR

Model Answer

1)  (x−3)² + (y−4)² = 25 (or x²+y²−6x−8y=0)

Explanation

1)  Standard form (x−h)²+(y−k)²=r² with (h,k)=(3,4), r=

5. Expand: x²−6x+9+y²−8y+16=25 ⇒ x²+y²−6x−8y=0.

PYQ2. For the circle x² + y² + Dx + Ey + F = 0, write the centre and the condition for a real circle.

1 mark(s) · SA · Sample QP 2024 · Q3(i) concept

Model Answer

1)  Centre (−D/2, −E/2); real iff D²+E²−4F ≥ 0 (radius ½√(D²+E²−4F))

Explanation

1)  Completing the square yields centre and radius formulae from the general equation of a circle.

PYQ3. Find the centre and radius of the circle x² + y² − 4x + 6y − 3 = 0.

2 mark(s) · SA · Board-style (from papers on circle centre)

Model Answer

1)  Centre (2, −3); radius 4

Explanation

1)  D=−4, E=6, F=−

3. Centre (2,−3). r=½√(16+36+12)=½√64=4.

PYQ4. Find the equation of the circle with diameter endpoints A(1, 2) and B(5, −2).

2 mark(s) · SA · Board-style diameter form

Model Answer

1)  x² + y² − 6x + 1 = 0

Explanation

1)  Diameter form (x−1)(x−5)+(y−2)(y+2)=0

2)  x²−6x+5+y²−4=0

3)  x²+y²−6x+1=0.

Problem Solving — L15 Circles

10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.

Question 1 of 10Standard form

Find the equation of the circle with centre (3, −2) and radius 5.

(x−h)²+(y−k)²=r²

1)  Substitute h=3, k=−2, r=5 into (x−h)²+(y−k)²=r².

2)  (x−3)² + (y+2)² = 25.

Answer:  (x−3)² + (y+2)² = 25

Formula used

(x−h)²+(y−k)²=r²

Textbook formal language

Standard circle equation with given centre and radius.

Easy language (same calculation)

Centre (3,−2), radius 5 → square the shifts and set equal to 25.

Why this formula

Expand only if asked for general form x²+y²+Dx+Ey+F=0.

Exam tip

y−(−2)=y+2.

Common mistakes

  • Writing (y−2) instead of (y+2)
  • Using r not r²
Question 2 of 10General form

Find the centre and radius of x² + y² − 4x + 6y − 3 = 0.

Centre (−D/2,−E/2), r=½√(D²+E²−4F)

1)  D=−4, E=6, F=−3.

2)  Centre = (4/2, −6/2) = (2, −3).

3)  r = ½√(16+36+12) = ½√64 = ½·8 = 4.

Answer:  Centre (2, −3); radius 4

Formula used

Centre (−D/2,−E/2), r=½√(D²+E²−4F)

Textbook formal language

Completing the square or using centre/radius formulae from the general equation.

Easy language (same calculation)

Half the opposite of linear coeffs for centre; then formula for r.

Why this formula

Need D²+E²−4F ≥ 0 for a real circle.

Exam tip

r = √(g²+f²−c) if equation is x²+y²+2gx+2fy+c=0 (equivalent form).

Common mistakes

  • Sign error on centre
  • Forgetting the 4F term
Question 3 of 10Diameter

Find the equation of the circle with diameter endpoints A(1, 2) and B(5, −2).

(x−x₁)(x−x₂)+(y−y₁)(y−y₂)=0

1)  (x−1)(x−5) + (y−2)(y+2) = 0.

2)  x² − 6x + 5 + y² − 4 = 0.

3)  x² + y² − 6x + 1 = 0.

Answer:  x² + y² − 6x + 1 = 0

Formula used

(x−x₁)(x−x₂)+(y−y₁)(y−y₂)=0

Textbook formal language

Diameter form of the circle with given endpoints of a diameter.

Easy language (same calculation)

Use the product form for diameter AB, expand, simplify.

Why this formula

Angle in a semicircle is 90° — geometric basis of the formula.

Exam tip

Expand carefully: (y−2)(y+2)=y²−4.

Common mistakes

  • Missing the constant terms
  • Using mid-point only without radius
Question 4 of 10Point position

Does the point (1, 1) lie inside, on, or outside the circle x² + y² = 4?

Compare OP² with r²

1)  For (1,1): 1²+1²=2.

2)  r²=4; since 2 < 4, the point is inside the circle.

Answer:  Inside (because 1+1=2 < 4)

Formula used

Compare OP² with r²

Textbook formal language

Compare the square of the distance from centre with r².

Easy language (same calculation)

1+1 is less than 4, so inside.

Why this formula

On the circle iff x²+y²=r² for centre origin.

Exam tip

Do not compare with r unless you take square roots consistently.

Common mistakes

  • Saying outside when 2<4
  • Using |x|+|y|
Question 5 of 10Tangent condition

Show that the line 3x + 4y − 10 = 0 is tangent to the circle x² + y² = 4.

d = r for tangency

1)  Centre (0,0), r=2.

2)  Distance from centre to line = |−10|/√(9+16) = 10/5 = 2.

3)  d = r = 2 ⇒ the line is tangent to the circle.

Answer:  Tangent (d = r = 2)

Formula used

d = r for tangency

Textbook formal language

A line is tangent to a circle iff the perpendicular distance from the centre equals the radius.

Easy language (same calculation)

Distance from origin to the line is 2, same as radius.

Why this formula

If d<r secant; d>r no real intersection.

Exam tip

Use absolute value in the distance formula.

Common mistakes

  • Comparing d with r²
  • Wrong √(A²+B²)
Question 6 of 10Expand

Expand (x−1)²+(y+2)²=9 into the form x²+y²+Dx+Ey+F=0 and state D,E,F.

(x−h)²+(y−k)²=r² → general form

1)  x²−2x+1 + y²+4y+4 = 9.

2)  x² + y² − 2x + 4y + 5 − 9 = 0.

3)  x² + y² − 2x + 4y − 4 = 0 ⇒ D=−2, E=4, F=−4.

Answer:  x²+y²−2x+4y−4=0 (D=−2, E=4, F=−4)

Formula used

(x−h)²+(y−k)²=r² → general form

Textbook formal language

Expanding the standard form yields the general circle equation with the stated coefficients.

Easy language (same calculation)

Open the brackets, bring 9 to the left, collect terms.

Why this formula

Centre is still (1,−2), radius 3.

Exam tip

Constant term is 1+4−9=−4.

Common mistakes

  • Losing the constant
  • Wrong sign on 4y
Question 7 of 10Standard circle

Write the equation of the circle with centre (−1, 4) and radius 5.

(x−h)²+(y−k)²=r²

1)  (x + 1)² + (y − 4)² = 25.

Answer:  (x+1)² + (y−4)² = 25

Formula used

(x−h)²+(y−k)²=r²

Textbook formal language

Standard form uses centre and r².

Easy language (same calculation)

Shift left 1, up 4, radius 5.

Why this formula

Expand only if general form is required.

Exam tip

Check each algebraic step carefully.

Common mistakes

  • r=5 not r²=25
  • Centre sign errors
Question 8 of 10General form

Find the centre and radius of x² + y² − 4x + 6y − 3 = 0.

Centre (−g,−f), r=√(g²+f²−c)

1)  2g=−4 ⇒ g=−2; 2f=6 ⇒ f=3; c=−3.

2)  Centre (−g,−f)=(2,−3).

3)  r = √(g²+f²−c)=√(4+9−(−3))=√16=4.

4)  Alternatively complete square: (x−2)²+(y+3)²=16.

Answer:  Centre (2, −3); radius 4

Formula used

Centre (−g,−f), r=√(g²+f²−c)

Textbook formal language

Completing the square or the (−g,−f) rule.

Easy language (same calculation)

Centre opposite half of linear terms; r from formula.

Why this formula

Need g²+f²−c > 0 for a real circle.

Exam tip

Check each algebraic step carefully.

Common mistakes

  • Centre (−2,3)
  • r² instead of r
Question 9 of 10On circle

Does the point (3, 4) lie on the circle x² + y² = 25?

Substitute into S=0

1)  3² + 4² = 9 + 16 = 25.

2)  Yes — the point satisfies the equation.

Answer:  Yes

Formula used

Substitute into S=0

Textbook formal language

A point lies on the circle iff it satisfies the equation.

Easy language (same calculation)

3-4-5 right triangle from origin.

Why this formula

Equality means on; < r² inside (for origin-centred).

Exam tip

Check each algebraic step carefully.

Common mistakes

  • Saying no
  • Using distance wrongly
Question 10 of 10Tangent

Find the equation of the tangent to x² + y² = 25 at the point (3, 4).

xx₁ + yy₁ = r² for x²+y²=r²

1)  Tangent: x·3 + y·4 = 25 ⇒ 3x + 4y = 25.

Answer:  3x + 4y = 25

Formula used

xx₁ + yy₁ = r² for x²+y²=r²

Textbook formal language

For the circle x²+y²=r² the tangent at (x₁,y₁) is xx₁+yy₁=r².

Easy language (same calculation)

Replace one x by 3 and one y by 4.

Why this formula

Radius to (3,4) is perpendicular to this tangent.

Exam tip

Check each algebraic step carefully.

Common mistakes

  • 3x+4y=5
  • Point not checked on circle