Equation of a Circle (AQA A-Level Mathematics): Revision Notes

A circle centred at the origin with radius r has the equation: $x^2 + y^2 = r^2$

If we perform an "inside transformation" to both $x$ and $y$, i.e.,

$(x + a)^2 + (y + b)^2 = r^2$

we translate $(-a, -b)$ so the circle is now centred at $(-a, -b)$.

A circle with the equation $(x + a)^2 + (y + b)^2 = r^2$ has radius r and centre $(-a, -b)$.

Example: Find the centre and radius of the circle with the equation

$x^2 + y^2 + 3x - 2y + 1 = 0$

Problem:

This looks nothing like the equation of a circle given above.

Solution:

Complete the square for $x$ and for $y$.

1. Start with the given equation: $x^2 + 3x + y^2 - 2y + 1 = 0$

2. Complete the square for $x$: $x^2 + 3x = \left( x + \frac{3}{2} \right)^2 - \frac{9}{4}$

3. Complete the square for $y$: $y^2 - 2y = (y - 1)^2 - 1$

4. Substitute these into the equation: $\left( x + \frac{3}{2} \right)^2 - \frac{9}{4} + (y - 1)^2 - 1 + 1 = 0$

5. Simplify and rearrange: $\left( x + \frac{3}{2} \right)^2 + (y - 1)^2 = \frac{9}{4}$

6. Identify the centre and radius: $\text{Center} = \left( -\frac{3}{2}, 1 \right)$

$\text{Radius} = \sqrt{\frac{9}{4}} = \frac{3}{2}$

Final Result

• Centre: $\left( -\frac{3}{2}, 1 \right)$
• Radius: $\frac{3}{2}$

Example: Write in the form $x^2 + y^2 + axe + by + c = 0$ the equation of the circle with centre $(2, 4)$ and radius $7$.

7. Equation of the circle: $(x - 2)^2 + (y - 4)^2 = 49$

• Centre: $(2, 4)$
• Radius: $r = \sqrt{49} = 7$

8. Expand and rearrange: $(x - 2)^2 + (y - 4)^2 = 49$

$x^2 - 4x + 4 + y^2 - 8y + 16 = 49$

$x^2 + y^2 - 4x - 8y + 20 - 49 = 0$

$x^2 + y^2 - 4x - 8y - 29 = 0$

For any circle question, draw a diagram.

Example: A circle has centre $(4, -3)$ and passes through $(10, 3)$. Find its equation.

1. Find the radius: $r = \sqrt{(10 - 4)^2 + (3 + 3)^2}$

$r = \sqrt{6^2 + 6^2}$

$r = \sqrt{36 + 36}$

$r = \sqrt{72}$

$r = 6\sqrt{2}$

9. Equation of the circle: $(x - 4)^2 + (y + 3)^2 = 72$

Example: Q3 (Jan 2010, Q8)

A circle has the equation $x^2 + y^2 + 6x - 4y - 4 = 0$.

(i) Find the centre and radius of the circle.

1. Complete the square: $x^2 + 6x + y^2 - 4y = 4$

$(x + 3)^2 - 9 + (y - 2)^2 - 4 = 4$

$(x + 3)^2 + (y - 2)^2 = 17$

2. Identify the centre and radius: $\text{Center} = (-3, 2)$

$\text{Radius} = \sqrt{17}$

(ii) Find the coordinates of the points where the circle meets the line with equation $y = 3x + 4$.

1. Substitute $y = 3x + 4$ into the circle equation: $x^2 + y^2 + 6x - 4y - 4 = 0$ $x^2 + (3x + 4)^2 + 6x - 4(3x + 4) - 4 = 0$

$x^2 + 9x^2 + 24x + 16 + 6x - 12x - 16 - 4 = 0$

$10x^2 + 18x = 0$

$x(10x + 18) = 0$

$x = 0$

$y = 3(0) + 4 = 4$

2. Find the $y$**-**coordinate for the other solution: $x = -\frac{18}{10}$

$x = -\frac{9}{5}$

$y = 3\left(-\frac{9}{5}\right) + 4$

$y = -\frac{27}{5} + \frac{20}{5}$

$y = -\frac{7}{5}$

3. Points of intersection: $(0, 4)$

$\left(-\frac{9}{5}, -\frac{7}{5}\right)$

4. Solve the quadratic equation: $10x^2 + 18x - 4 = 0$

• Using the quadratic formula or a calculator: $x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 10 \cdot (-4)}}{2 \cdot 10}$

$x = \frac{-18 \pm \sqrt{324 + 160}}{20}$

$x = \frac{-18 \pm \sqrt{484}}{20}$

$x = \frac{-18 \pm 22}{20}$

$x = \frac{4}{20} = \frac{1}{5}$

$x = \frac{-40}{20} = -2$

5. Find the corresponding $y$-values:

• For $x = \frac{1}{5}$: $y = 3\left(\frac{1}{5}\right) + 4 = \frac{3}{5} + 4 = \frac{3}{5} + \frac{20}{5} = \frac{23}{5} \left( \therefore \frac{1}{5}, \frac{23}{5} \right)$

• For $x = -2$: $y = 3(-2) + 4 = -6 + 4 = -2$ $(-2, -2)$

Final Points of Intersection:

$\left( \frac{1}{5}, \frac{23}{5} \right)$

$(-2, -2)$

These are the points where the circle meets the line $y = 3x + 4$.