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Equations of lines and circles

Level 6 · Ages 16–17

Quick recall· Let's wake up what you already know before we start.

Recall: what is the slope of the line through (0, 2) and (3, 11)?

The whole lesson

Read it again any time, including while you work through the exercises.

A curve's equation is a membership test: a point (x, y) is on the curve if and only if its coordinates satisfy the equation. This viewpoint turns geometry into computation, one of the most powerful ideas in mathematics.

For lines, you know the form y = mx + p. There is also the Cartesian form ax + by + c = 0, more general since it also covers vertical lines. To recover the slope of a line given in that form (with b nonzero), solve for y: the coefficient of x that appears is the slope.

For circles, everything flows from the definition: the circle with centre C(a, b) and radius r is the set of points at distance r from C. With the distance formula (Pythagoras!), this gives the equation (x - a)² + (y - b)² = r². Mind the signs: (x - 2)² corresponds to a centre with x-coordinate +2, and the right side is r², not r!

Sometimes a circle hides in expanded form, like x² + y² - 4x + 6y - 3 = 0. To unmask it, complete the squares: group the x terms, the y terms, and reveal (x - a)² and (y - b)². The centre and radius can then be read off directly.

With these equations you can test whether a point is on a circle, inside or outside: compare its distance to the centre with the radius. A true geometric GPS!

The worked example, step by step

Write the equation of the circle with centre C(2, -3) and radius 4.

  1. (x - a)² + (y - b)² = r²

    I start from the general form with centre (a, b) and radius r. It says: the distance to the centre equals r.

  2. (x - 2)² + (y - (-3))² = 4²

    I substitute: a = 2, b = -3, r = 4. Watch the sign trap: y - (-3) becomes y + 3.

  3. (x - 2)² + (y + 3)² = 16

    I simplify: the right side is r² = 16, not 4! That is the circle's equation.

Result: (x - 2)² + (y + 3)² = 16.

The equation x² + y² - 4x + 6y - 3 = 0 defines a circle. Find its centre and radius.

  1. x² - 4x = (x - 2)² - 4

    I complete the square in x: x² - 4x is (x - 2)² minus the extra 4.

  2. y² + 6y = (y + 3)² - 9

    Same work in y: y² + 6y is (y + 3)² minus 9.

  3. (x - 2)² + (y + 3)² - 4 - 9 - 3 = 0

    I substitute into the equation and gather the constants: -4 - 9 - 3 = -16.

  4. (x - 2)² + (y + 3)² = 16

    I recognise the circle form: centre (2, -3) and r² = 16, so radius 4. The signs flip when reading the centre!

Result: Centre (2, -3), radius 4.