Strand 2 — Geometry & Trigonometry

Co-ordinate Geometry of the Circle

5th Year · 6th Year (Leaving Cert)

✓By the end of this lesson students will be able to derive and apply the equation of a circle with a given centre and radius.
✓By the end of this lesson students will be able to convert between the standard form and the general form of a circle's equation, and determine its centre and radius.
✓By the end of this lesson students will be able to find the equation of a tangent to a circle at a given point on the circle.
✓By the end of this lesson students will be able to solve problems involving the intersection of a line and a circle, or two circles.

Key concepts

Equation of a Circle (Standard Form)

The equation of a circle with centre (h, k) and radius r is derived from the distance formula. Any point (x, y) on the circle is a distance r from the centre (h, k).

(x - h)² + (y - k)² = r²

Equation of a Circle (Centre at Origin)

This is a special case of the standard form where the centre (h, k) is (0, 0).

x² + y² = r²

General Equation of a Circle

Expanding the standard form (x - h)² + (y - k)² = r² gives a general quadratic equation. Comparing coefficients allows us to find the centre and radius from this form.

x² + y² + 2gx + 2fy + c = 0 Centre: (-g, -f) Radius: √(g² + f² - c) For a real circle, g² + f² - c > 0.

Position of a Point Relative to a Circle

For a circle S: (x - h)² + (y - k)² - r² = 0, or S: x² + y² + 2gx + 2fy + c = 0, and a point P(x₁, y₁):

S(x₁, y₁) = 0 if P is on the circle. S(x₁, y₁) < 0 if P is inside the circle. S(x₁, y₁) > 0 if P is outside the circle.

Equation of a Tangent to a Circle at a Point (x₁, y₁) on the Circle

The tangent to a circle at a point (x₁, y₁) on the circle is perpendicular to the radius drawn to that point. A general formula can be used for any form of the circle's equation.

For x² + y² + 2gx + 2fy + c = 0, the tangent at (x₁, y₁) is: xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0

Intersection of a Line and a Circle

To find the points of intersection, substitute the equation of the line into the equation of the circle. This will result in a quadratic equation. The discriminant (Δ = b² - 4ac) of this quadratic determines the nature of the intersection:

Δ > 0: Two distinct points of intersection (the line is a secant). Δ = 0: One point of intersection (the line is a tangent). Δ < 0: No points of intersection.

Key facts to remember

1The standard form of a circle's equation is (x - h)² + (y - k)² = r², where (h, k) is the centre and r is the radius.
2The general form of a circle's equation is x² + y² + 2gx + 2fy + c = 0. Its centre is (-g, -f) and its radius is √(g² + f² - c).
3For a real circle to exist, the radius squared must be positive: g² + f² - c > 0.
4The equation of the tangent to the circle x² + y² + 2gx + 2fy + c = 0 at a point (x₁, y₁) on the circle is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0.
5The radius drawn to the point of tangency is always perpendicular to the tangent line.
6To find the intersection of a line and a circle, substitute the line's equation into the circle's equation and solve the resulting quadratic. The discriminant determines the number of intersection points.

Worked examples

Example 1

Find the equation of the circle with centre C(2, -3) and radius 5. Express your answer in both standard form and general form.

IThe centre is (h, k) = (2, -3) and the radius is r = 5.

IISubstitute these values into the standard form equation: (x - h)² + (y - k)² = r²

III(x - 2)² + (y - (-3))² = 5²

IV(x - 2)² + (y + 3)² = 25 (This is the standard form)

VTo convert to general form, expand the brackets:

VI(x² - 4x + 4) + (y² + 6y + 9) = 25

VIIx² - 4x + 4 + y² + 6y + 9 - 25 = 0

VIIIx² + y² - 4x + 6y - 12 = 0 (This is the general form)

Answer

Standard form: (x - 2)² + (y + 3)² = 25 General form: x² + y² - 4x + 6y - 12 = 0

Example 2

A circle has the equation x² + y² - 6x + 4y - 12 = 0. Find its centre and radius. Determine if the point P(7, 1) lies on, inside, or outside the circle.

ICompare the given equation x² + y² - 6x + 4y - 12 = 0 with the general form x² + y² + 2gx + 2fy + c = 0.

IIFrom comparison: 2g = -6 => g = -3

III2f = 4 => f = 2

IVc = -12

VThe centre of the circle is (-g, -f) = (-(-3), -2) = (3, -2).

VIThe radius of the circle is r = √(g² + f² - c)

VIIr = √((-3)² + (2)² - (-12))

VIIIr = √(9 + 4 + 12)

9r = √25 = 5

10To determine the position of point P(7, 1), substitute its coordinates into the circle's equation (let S(x, y) = x² + y² - 6x + 4y - 12):

11S(7, 1) = (7)² + (1)² - 6(7) + 4(1) - 12

12S(7, 1) = 49 + 1 - 42 + 4 - 12

13S(7, 1) = 54 - 54 = 0

14Since S(7, 1) = 0, the point P(7, 1) lies on the circle.

Answer

Centre: (3, -2) Radius: 5 The point P(7, 1) lies on the circle.

Example 3

Find the equation of the tangent to the circle x² + y² + 4x - 2y - 20 = 0 at the point A(3, 5).

IFirst, verify that A(3, 5) is on the circle:

II(3)² + (5)² + 4(3) - 2(5) - 20 = 9 + 25 + 12 - 10 - 20 = 46 - 40 = 6 ≠ 0. (Correction: The point (3,5) is NOT on the circle. Let's re-evaluate the problem or assume it is on the circle for the purpose of the example, or pick a point that IS on the circle. Let's pick a point that is on the circle for a valid example. For x² + y² + 4x - 2y - 20 = 0, let's test (2,4): 4+16+8-8-20 = 0. So (2,4) is on the circle. I will use (2,4) instead of (3,5).)

IIIRevised Problem: Find the equation of the tangent to the circle x² + y² + 4x - 2y - 20 = 0 at the point A(2, 4).

IVVerify A(2, 4) is on the circle: (2)² + (4)² + 4(2) - 2(4) - 20 = 4 + 16 + 8 - 8 - 20 = 28 - 28 = 0. Yes, it is on the circle.

VCompare the circle equation x² + y² + 4x - 2y - 20 = 0 with x² + y² + 2gx + 2fy + c = 0.

VIWe have: 2g = 4 => g = 2

VII2f = -2 => f = -1

VIIIc = -20

9The point of tangency is (x₁, y₁) = (2, 4).

10Use the formula for the tangent at (x₁, y₁): xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0

11Substitute the values: x(2) + y(4) + 2(x + 2) + (-1)(y + 4) - 20 = 0

122x + 4y + 2x + 4 - y - 4 - 20 = 0

13Combine like terms:

144x + 3y - 20 = 0

Answer

The equation of the tangent is 4x + 3y - 20 = 0.

Always verify that the given point lies on the circle before applying the tangent formula for a point on the circle.

Example 4

Find the points of intersection of the line L: y = x + 1 and the circle K: x² + y² = 5.

ISubstitute the expression for y from the line equation (y = x + 1) into the circle equation (x² + y² = 5).

IIx² + (x + 1)² = 5

IIIExpand (x + 1)²:

IVx² + (x² + 2x + 1) = 5

VCombine like terms and rearrange into a quadratic equation:

VI2x² + 2x + 1 - 5 = 0

VII2x² + 2x - 4 = 0

VIIIDivide by 2 to simplify:

9x² + x - 2 = 0

10Factorise the quadratic equation:

11(x + 2)(x - 1) = 0

12This gives two possible values for x: x = -2 or x = 1.

13Substitute these x values back into the line equation y = x + 1 to find the corresponding y values:

14For x = -2: y = (-2) + 1 = -1. Point 1: (-2, -1)

15For x = 1: y = (1) + 1 = 2. Point 2: (1, 2)

16The points of intersection are (-2, -1) and (1, 2).

Answer

The points of intersection are (-2, -1) and (1, 2).

If the quadratic equation had no real solutions (discriminant < 0), there would be no points of intersection. If it had one repeated solution (discriminant = 0), the line would be tangent to the circle.

Common mistakes

✗Incorrectly identifying the signs of g and f when finding the centre from the general equation (e.g., using (g, f) instead of (-g, -f)).
✗Forgetting to square the radius (r) when writing the equation of a circle, or forgetting to take the square root when calculating the radius from g² + f² - c.
✗Algebraic errors when expanding (x - h)² or (y - k)² or when solving quadratic equations.
✗Applying the tangent formula for a point that is not actually on the circle.
✗Confusing the slope of the radius with the slope of the tangent; remember they are negative reciprocals of each other.

Exam tips

★Always draw a clear sketch of the problem, especially for tangent questions, to help visualise the geometry and check your answer.
★Double-check all signs and calculations, particularly when dealing with negative numbers and squares.
★When finding the equation of a tangent at a point, always verify that the given point lies on the circle first.
★Be proficient in both the formulaic method (using the general tangent formula) and the geometric method (using perpendicular slopes) for finding tangent equations, as one might be more suitable depending on the problem.
★Practise solving quadratic equations accurately and efficiently, as they are fundamental to intersection problems.

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