MATHEMATICS As LEVEL(FORM FIVE) NOTES – CIRCLE

CIRCLE

Is the locus shown on xy-plane so that it is always a constant distance, i.e., radius, from a fixed point, i.e., Centre.

Where the plural of radius is radii.

GENERAL EQUATION OF CIRCLE

Consider the figure below on X–Y plane.

Square both sides:

r² = (x – a)² + (y – b)²

Hence,

r² = (x – a)² + (y – b)²

The general formula for the equation of the circle:

By extending the above formula, we get the general equation of the circle:

(x – a)² + (y – b)² = r²

Expanding:

x² – 2ax + a² + y² – 2by + b² = r²

Rearranged:

x² + y² – 2ax – 2by + a² + b² – r² = 0

Let -a = g and -b = f, then:

x² + y² + 2gx + 2fy + c = 0

Therefore,

x² + y² + 2gx + 2fy + c = 0

Also, the coordinate of the center of the curve is (-g, -f).

1. To find the coordinate of the centre and radius of the circle, make sure the coefficient of x² and y² is equal to 1.
2. The centre of the curve is (-½ coefficient of x, -½ coefficient of y).
3. If the circle passes through the origin, then c = 0.
4. If the centre lies on the X-axis, then f = 0.
5. If the centre lies on the Y-axis, then g = 0.
6. If the coordinate of the centre lies at the origin (0,0), the general equation is x² + y² = r².

DETERMINATION OF CENTRE AND RADIUS OF THE CIRCLE

Mostly done by using two common methods:

1. By using the general equation of the circle.
2. By using the completing the square method.

I. By using the general equation of the circle

Compare the given equation with the general form x² + y² + 2gx + 2fy + c = 0 to find g, f, and c.

Coordinate of centre is (-g, -f).

II. By completing the square method

This method expresses the equation in the form (x – a)² + (y – b)² = r².

Where:

• Coordinate of the centre is taken under opposite signs of the parts of x and y.
• Radius is the square root of the constant term on the right side.

Example: (x – 2)² + (y + 3)² = 9

Centre: (2, -3)

Radius: 3

EXAMPLES

1. Determine the coordinate of the centre and radius of the following:
   • (a) x² + y² – 4x – 6y – 12 = 0
   • (b) 4x² + 4y² – 20x – 4y + 16 = 0
2. Write down the standard equation of the circle with centre at the origin and radius 5 units.
3. Find the equation of the circle passing through (2, 1) and centre at (-3, -4).

EQUATION OF THE CIRCLE WITH TWO POINTS AS THE END OF DIAMETER

Let AB be the diameter of the circle with coordinates A (x₁, y₁) and B (x₂, y₂). The general equation is:

(x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0

Proof considers the figure below with two points as diameter.

Required equation of the circle, let C be the angle formed in the semicircle.

Where ∠ABC = 90°

Since ∠ACB = 90º, then

Slope AB × Slope BC = -1

But

Slope AC = -1 and Slope BC = 1

EQUATION OF THE CIRCLE WITH THREE POINTS ON THE CIRCLE

Let us consider a circle with three points on the circle such as A, B, and C as shown below.

The equation of the circle above can be calculated by:

1. Using the general equation of the circle.
2. Using the distance formula.

I. By general equation

Refer to the general equation x² + y² + 2gx + 2fy + c = 0. Substitute the three points to form three equations in terms of g, f, and c. Solve to find the standard equation.

II. By distance formula

Use the distance formula to form two equations in terms of the centre coordinates (a, b) from the centre to points on the circle. Solve to find the centre and radius.

Example:

Where AO = BC and BO = OC.

EQUATION OF CIRCLE WITH TWO POINTS AND LINE PASSING THROUGH THE CENTRE

Let two points be A (x₁, y₁) and B (x₂, y₂) and line L₁: Px + Qy + C = 0 passing through the centre.

We need to find the equation of the circle.

Important steps

1. Since the line passes through the centre, the centre coordinates satisfy the line equation Px + Qy + C = 0.
2. Form the second equation using the distance formula in terms of a’ and b’ (coordinates of the centre).
3. By solving the equations, find the centre.

EXAMPLES

1. Find the equation of the circle on the line forming the points (-1, 2) and (-3, 5) as the diameter.
2. Find the length of tangent from (1, 2) to the circle x² + y² – 4x – 2y + 4 = 0.
3. Find the length of tangent from the origin to the circle x² + y² – 10x + 2y + 13 = 0.
4. Find the equation of tangent from (-1, 7) to the circle x² + y² = 5.
5. Find the equation of tangent to the circle x² + y² = 16 if the slope of tangent is 2.
6. Examine whether the point (2, 3) lies inside or outside the circle.
7. Discuss the position of points (1, 2) and (6, 0) with respect to the circle x² + y² – 4x – 2y – 11 = 0.

EQUATION OF TANGENT TO THE CIRCLE WITH COORDINATE OF CENTRE AND EQUATION OF TANGENT LINE

Let the centre be C(a, b) and the tangent line be px + qy – c = 0.

Our intention is to find the equation of the circle.

Important steps

1. Determine the radius as the shortest distance from the centre to the line.
2. Using the radius and centre, find the equation of the circle.

EQUATION OF TANGENT TO THE CIRCLE AT A GIVEN POINT

Our intention is to find the equation of the tangent. Let us find the slope by calculus method.

PROBLEMS

1. Find the equation of the circle with centre (3, -4) and tangent to line 3x + 4y + 3 = 0.
2. Find the equation of the circle whose centre is 9 units to the right of the Y-axis, radius 2 units, and touches the line 2x + y – 10 = 0.
3. Find the equation of tangent to the circle x² + y² = 18 at point (7, -3).
4. Find the equation of tangent for the circle through point (2, -4) given that x² + y² – 2x – 4y + 1 = 0.
5. Show that the point (7, -5) lies on the circle x² + y² – 6x + 4y – 12 = 0. Find the equation of tangent to the circle at the point.

THE CONDITION OF A CERTAIN LINE TO BE TANGENT TO THE CIRCLE

For y = Mx + c to be tangent to the circle x² + y² + 2gx + 2fy + c = 0, the discriminant condition must be satisfied:

b² = 4ac

Examples:

1. Find the value of K if 12x + 5y + K = 0 is tangent to the circle x² + y² – 6x -10y + 9 = 0.
2. Show that 5x + 12y – 4 = 0 touches the circle x² + y² – 6x + 4y + 12 = 0.

DISTANCE OF TANGENT FROM EXTERNAL POINT TO THE CIRCLE

Consider the figure below.

By Pythagoras theorem:

CP² + PT² = CT²

But CP = distance from centre to external point, CT = radius, PT = length of tangent.

ORTHOGONAL CIRCLES

This circle is said to be orthogonal only if their radii are perpendicular to each other.

Simply, their radii meet at right angle.

Means that circle I and II are orthogonal.

The condition for orthogonal circles is explained by the Pythagoras theorem:

r₁² + r₂² = d²

Where C₁C₂ = distance between centres, r₁ and r₂ are radii of the circles.

INTERSECTION OF TWO CIRCLES

Intersection of two circles may be classified into three cases:

1. Common tangent: line at which two circles intersect at a single point.
2. Common chord: line at which two circles intersect at two distinct points.
3. Line of separation: line between two circles with no points of intersection.

How to find common tangent, chord or line of separation?

This is done by subtracting the equation of one circle from another, i.e., C₁ – C₂ or C₂ – C₁.

The difference between two equations of circles represents the common tangent, chord, or line of separation.

Where the point of intersection of either common tangent or common chord is obtained by solving them simultaneously under the following:

1. If b² = 4ac, the line is a common tangent.
2. If b² > 4ac, the line is a common chord.
3. If b² < 4ac, the line is a line of separation.

CONCYCLIC CIRCLE

These are two circles which pass through the same coordinate of centre but vary in radius.

If four points are concyclic, then by forming the equation of the concyclic circle using three points, the fourth point should satisfy the equation.

EXAMPLES

1. Show that the part of the line 3y = x + 5 is a chord of the circle x² + y² – 6x – 2y – 15 = 0.
2. Find the equation of common chord and the intersection point of the circles x² + y² + 6x – 3y + 4 = 0 and 2x² + 2y² – 3x – 9y + 2 = 0.
3. Find the length of the chord of the circle x² + y² – 2x – 4y – 5 = 0, whose midpoint is (2, 3).
4. Show that the circles x² + y² – 4x + 6y – 10 = 0 and x² + y² – 10x + 6y + 14 = 0 intersect.
5. Find the equation of the circle concentric with the circle x² + y² + 4x + 6y + 11 = 0 and passing through (5, 4).
6. Find the equation of the circle which passes through the centre, x² + y² + 8x + 10y – 7 = 0, and is concentric with the circle 2x² + 2y² – 8x – 12y – 9 = 0.
7. Find the equation of the circle concentric with the circle 2x² + 2y² + 8x + 10y – 39 = 0 and having its area equal to 16.