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Applied Mathematics — Second Secondary
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Applied Mathematics — Second Secondary
Unit 1: Statics
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Forces
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Forces — Lesson quiz
6 questions
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Resolution of Forces
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Resolution of Forces — Lesson quiz
6 questions
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Resultant of Coplanar Concurrent Forces
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Resultant of Coplanar Concurrent Forces — Lesson quiz
6 questions
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Equilibrium of a Rigid Body under Coplanar Forces
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Equilibrium of a Rigid Body under Coplanar Forces — Lesson quiz
6 questions
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Equilibrium on a Rough Horizontal Plane
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Equilibrium on a Rough Horizontal Plane — Lesson quiz
6 questions
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Unit 2: Solid and Analytic Geometry
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Lines and Planes in Space
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Lines and Planes in Space — Lesson quiz
6 questions
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The Pyramid and the Cone
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The Pyramid and the Cone — Lesson quiz
6 questions
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Lateral and Total Area of Pyramid and Cone
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Lateral and Total Area of Pyramid and Cone — Lesson quiz
6 questions
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Volume of the Right Pyramid and Cone
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Volume of the Right Pyramid and Cone — Lesson quiz
6 questions
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Equation of a Circle
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Equation of a circle with given centre and radius
The general form of the equation of a circle
Position of a point relative to a circle
Worked examples
Common mistakes
Lesson summary
Equation of a Circle — Lesson quiz
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Text lesson
Equation of a Circle
Applied Mathematics — Second Secondary
Equation of a Circle
Explanation
Worked examples
Worked examples
Example (1): Writing the equation from the centre and radius
Find the equation of the circle with centre (2, −3) and radius 4, then check whether the point (2, 1) lies on it.
Solution:
(x − 2)² + (y + 3)² = 4² = 16.
(2 − 2)² + (1 + 3)² = 0 + 16 = 16.
Both sides are equal, so the point (2, 1) lies on the circle.
Example (2): Converting the general equation to standard form
Find the centre and the radius of the circle whose equation is x² + y² − 4x + 6y − 3 = 0.
Solution:
2g = −4, so g = −2; 2f = 6, so f = 3; and c = −3.
Centre = (−g, −f) = (2, −3).
r² = g² + f² − c = 4 + 9 + 3 = 16, so r = 4.
(x − 2)² + (y + 3)² = 16.
Example (3): A circle with a known diameter
Find the equation of the circle whose diameter has end points A(1, 2) and B(5, 6).
Solution:
((1 + 5) ÷ 2, (2 + 6) ÷ 2) = (3, 4).
AB = √((5 − 1)² + (6 − 2)²) = √32 = 4√2, so the radius is 2√2.
r² = 8, so the equation is (x − 3)² + (y − 4)² = 8.
(1 − 3)² + (2 − 4)² = 4 + 4 = 8.
Example (4): A circle centred at the origin, and the position of a point
Find the equation of the circle with centre at the origin that passes through (3, 4), then determine the position of the point (4, 4) relative to it.
Solution:
r² = 3² + 4² = 25, so the equation is x² + y² = 25.
4² + 4² = 32.
Since 32 > 25, the point (4, 4) lies outside the circle.
Example (5): A circle touching the y-axis
Find the equation of the circle with centre (−1, 2) that touches the y-axis.
Solution:
The distance from the centre to the y-axis is |−1| = 1. Since the circle touches the axis, this distance equals the radius.
The equation is (x + 1)² + (y − 2)² = 1.
Example (6): An equation that does not represent a real circle
Decide whether x² + y² + 2x − 4y + 6 = 0 represents a real circle.
Solution:
g = 1, f = −2, c = 6.
g² + f² − c = 1 + 4 − 6 = −1 < 0.
So the equation does not represent a real circle, because no point satisfies it.
Equation of a Circle · 4 / 6
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