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O-Level Additional Mathematics · Secondary 3 and 4

Coordinate Geometry of Circles

This topic extends coordinate geometry from straight lines to circles. You learn the equation of a circle in two forms, find the centre and radius from an equation, and solve problems where lines are tangents to a circle or meet it. Earlier skills come together here: the gradient and midpoint formulae, perpendicular gradients, completing the square and the discriminant. A typical question gives a few facts about a circle, asks you to build its equation, then asks for a tangent or a point of intersection.

What the syllabus expects

  • Use the equation of a circle with centre (a, b) and radius r, (x - a)² + (y - b)² = r²
  • Find the centre and radius from the general form x² + y² + 2gx + 2fy + c = 0
  • Use the facts that a tangent is perpendicular to the radius at the point of contact and that the perpendicular bisector of a chord passes through the centre
  • Find the equation of a tangent to a circle at a given point
  • Find the points where a line meets a circle and decide whether a line cuts, touches or misses a circle
  • Solve problems that combine circles with midpoints, gradients and lengths of line segments

The equation of a circle

A point (x, y) lies on a circle with centre (a, b) and radius r exactly when its distance from the centre is r. Squaring the distance formula gives (x - a)² + (y - b)² = r². For example, the circle with centre (3, -2) and radius 5 has equation (x - 3)² + (y + 2)² = 25. Watch the signs: the coordinates of the centre have the opposite signs to the numbers in the brackets. A circle with its centre at the origin has equation x² + y² = r². To check whether a point lies on, inside or outside a circle, substitute it into the left side and compare with r²: equal means on the circle, smaller means inside and larger means outside.

The general form

Expanding (x - a)² + (y - b)² = r² gives an equation of the form x² + y² + 2gx + 2fy + c = 0. Comparing the two forms, the centre is (-g, -f) and the radius is √(g² + f² - c). For example, x² + y² - 6x + 4y - 12 = 0 has 2g = -6 and 2f = 4, so g = -3 and f = 2. The centre is (3, -2) and the radius is √(9 + 4 + 12) = √25 = 5. You can also complete the square: (x - 3)² - 9 + (y + 2)² - 4 - 12 = 0, so (x - 3)² + (y + 2)² = 25. In a circle equation, the coefficients of x² and y² are equal and there is no xy term. If the equation starts with 2x² + 2y², divide through by 2 first.

Tangents and chords

Two geometric facts solve most circle problems. First, the tangent at a point P is perpendicular to the radius CP, where C is the centre. So if the gradient of CP is m, the gradient of the tangent is -1/m. Second, the perpendicular bisector of any chord passes through the centre. So if you know two points on a circle, the centre lies on the perpendicular bisector of the segment joining them. For example, for the circle (x - 3)² + (y + 2)² = 25, the point P(6, 2) is on the circle since 3² + 4² = 25. The gradient of CP is (2 - (-2))/(6 - 3) = 4/3, so the tangent at P has gradient -3/4 and equation y - 2 = -3/4(x - 6), which simplifies to 4y = -3x + 26.

Lines meeting circles

To find where a line meets a circle, substitute the equation of the line into the equation of the circle to get a quadratic in one variable. Two distinct real roots means the line cuts the circle at two points, a repeated root means the line is a tangent, and no real roots means the line misses the circle. This method also gives the coordinates of the points of intersection. For example, substituting y = x + 1 into x² + y² = 13 gives x² + (x + 1)² = 13, so 2x² + 2x - 12 = 0, which simplifies to x² + x - 6 = 0 and gives x = 2 or x = -3. The points are (2, 3) and (-3, -2). Each point can be checked in both equations.

Worked example

A circle passes through the points A(1, 4) and B(5, 2), and its centre lies on the line y = x - 1. Find the equation of the circle.

  1. The centre lies on the perpendicular bisector of the chord AB. The midpoint of AB is ((1 + 5)/2, (4 + 2)/2) = (3, 3).
  2. The gradient of AB is (2 - 4)/(5 - 1) = -1/2, so the perpendicular gradient is 2.
  3. The perpendicular bisector is y - 3 = 2(x - 3), which is y = 2x - 3.
  4. Solve with y = x - 1: 2x - 3 = x - 1, so x = 2 and y = 1. The centre is (2, 1).
  5. Radius squared = (1 - 2)² + (4 - 1)² = 1 + 9 = 10. Check with B: (5 - 2)² + (2 - 1)² = 9 + 1 = 10.
  6. The equation is (x - 2)² + (y - 1)² = 10, which expands to x² + y² - 4x - 2y - 5 = 0.

Answer: (x - 2)² + (y - 1)² = 10, or x² + y² - 4x - 2y - 5 = 0

Common mistakes

  • Reading the centre of (x + 4)² + (y - 1)² = 9 as (4, -1). The centre is (-4, 1), because the brackets are (x - a) and (y - b).
  • Giving the radius of (x + 4)² + (y - 1)² = 9 as 9. The right side is r², so the radius is 3.
  • Forgetting the constant c when finding the radius from the general form, or adding it instead of subtracting. The radius is √(g² + f² - c).
  • Using the gradient of the radius as the gradient of the tangent. The tangent is perpendicular to the radius, so its gradient is the negative reciprocal.
  • Reading off g and f when the coefficients of x² and y² are not 1. Divide the whole equation first, so 2x² + 2y² - 8x + 4y = 0 becomes x² + y² - 4x + 2y = 0.
  • Expanding (x + 1)² as x² + 1 after substituting a line into a circle. Expand every squared bracket in full: (x + 1)² = x² + 2x + 1.

Exam tips

  • Draw a quick sketch with the centre, the radius and any given points marked. It shows which fact to use, the perpendicular bisector or the tangent.
  • When a circle touches the x-axis, the radius equals the size of the y-coordinate of the centre, and when it touches the y-axis, the radius equals the size of the x-coordinate. Questions often give the radius this way.
  • After finding an equation, substitute one of the given points back in. It catches sign errors in the centre straight away.

Practice questions

1. Find the centre and radius of the circle x² + y² + 8x - 2y - 8 = 0.

Show answer

Centre (-4, 1) and radius 5. Here g = 4, f = -1 and c = -8, so the centre is (-g, -f) = (-4, 1) and the radius is √(16 + 1 + 8) = √25 = 5.

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

Show answer

3x + 4y = 25. The centre is (0, 0), so the radius to (3, 4) has gradient 4/3 and the tangent has gradient -3/4. Then y - 4 = -3/4(x - 3), so 4y - 16 = -3x + 9, which gives 3x + 4y = 25.

3. Find the range of values of k for which the line y = x + k meets the circle x² + y² = 8.

Show answer

-4 ≤ k ≤ 4. Substituting gives x² + (x + k)² = 8, so 2x² + 2kx + k² - 8 = 0. For the line to meet the circle, b² - 4ac ≥ 0: 4k² - 8(k² - 8) ≥ 0, which simplifies to 64 - 4k² ≥ 0, so k² ≤ 16. The line is a tangent when k = 4 or k = -4.

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