1.3.2 Equation of a Circle

In this topic we will learn how to:

• understand that the equation, $(x - a)^{2} + (y - b)^{2} = r^{2}$ represents the circle with centre $(a, b)$ and radius $r$
• use the expanded form $x^{2} + y^{2} + 2gx + 2fy + c = 0$
• use algebraic methods to solve problems involving lines and circles

Note: Knowledge of geometrical properties of a circle, such as circle theorems, is assumed.

The equation of a circle is written in the form,
$$(x - a)^{2} + (y - b)^{2} = r^{2}$$Where $(a, b)$ represents the coordinates of the centre of the circle, and $r$ represents the radius.

The equation of a circle can also be written as the expanded form,
$$x^{2} + y^{2} + 2gx + 2fy + c = 0$$Where $g$, $f$ and $c$ are constants.

To get from the expanded form to the generic form, we complete the square. See example number $1$ below.

Let’s look at some past paper questions.

1. The equation of the circle with centre $C$ is $x^{2} + y^{2} - 8x + 4y - 5 = 0$ (9709/12/M/J/20 number 11)

(a) Find the radius of the circle and coordinates of $C$.
$$x^{2} + y^{2} - 8x + 4y - 5 = 0$$We have been given the expanded equation of the circle. We must complete the square to get to the generic form.

Start by putting terms in $x$ together and terms in $y$ together,
$$\textcolor{#2192ff}{x^{2} - 8x} + \textcolor{#0f0}{y^{2} + 4y} - 5 = 0$$Take any constants to the right-hand side,
$$\textcolor{#2192ff}{x^{2} - 8x} + \textcolor{#0f0}{y^{2} + 4y} = 5$$Put one set of brackets on the terms in $x$ and one set on the terms in $y$,
$$(\textcolor{#2192ff}{x^{2} - 8x}) + (\textcolor{#0f0}{y^{2} + 4y}) = 5$$Complete the square for both the brackets, separately,
$$\textcolor{#2192ff}{(x - 4)^{2} - 16} + \textcolor{#0f0}{(y + 2)^{2} - 4} = 5$$Move all constants to the right-hand side,
$$(x - 4)^{2} + (y + 2)^{2} = 5 + 4 + 16$$Simplify,
$$(x - 4)^{2} + (y + 2)^{2} = 25$$$$(x - \textcolor{#2192ff}{4})^{2} + (y - \textcolor{#2192ff}{(-2)})^{2} = 25$$Now that we have the generic equation of the circle, we can read of the coordinates of $C$,
$$C(\textcolor{#2192ff}{4}, \textcolor{#2192ff}{-2})$$Let’s calculate the radius,
$$r^{2} = 25$$$$\sqrt{r^{2}} = \sqrt{25}$$$$r = 5$$Therefore, the final answer is,
$$C(4, -2)\ \ \ \ \ \ r = 5$$(b) The point $P(1, 2)$ lies on the circle. Show that the equation of the tangent to the circle at $P$ is $4y = 3x + 5$.

Let’s sketch a diagram of the problem,

To find the equation of the tangent, we need to find its gradient first. The tangent is perpendicular to $CP$, so let’s find the gradient of $CP$,
$$\textmd{gradient of CP} = \frac{2 + 2}{1 - 4}$$$$\textmd{gradient of CP} = -\frac{4}{3}$$Since $CP$ is perpendicular to the tangent,
$$m_{1} \times m_{2} = -1$$$$-\frac{4}{3} \times m_{2} = -1$$$$m_{2} = \frac{3}{4}$$Therefore, the gradient of the tangent is,
$$\frac{3}{4}$$Now let’s find the equation of the tangent, and remember that the tangent passes through $P$,
$$m = \frac{3}{4}\ \ \ \ \ \ P(1, 2)$$$$y = mx + c$$$$2 = \frac{3}{4}(1) + c$$$$2 = \frac{3}{4} + c$$$$c = 2 - \frac{3}{4}$$$$c = \frac{5}{4}$$$$y = \frac{3}{4}x + \frac{5}{4}$$Multiply through by $4$,
$$4y = 3x + 5$$Therefore, the final answer is,
$$4y = 3x + 5$$2. The coordinates of two points $A$ and $B$ are $(-7, 3)$ and $(5, 11)$ respectively. The perpendicular bisector of $AB$ has the equation $3x + 2y = 11$. A circle passes through $A$ and $B$ and its center lies on the line $12x - 5y = 70$. Find an equation of the circle. (9709/13/M/J/20 number 10)

Let’s sketch a diagram of the problem,

The perpendicular bisector and the line $\textcolor{#0f0}{12x - 5y = 70}$, intersect at the center of the circle, $C$. So let’s solve simultaneously to find the coordinates of C,
$$3x + 2y = 11\ \ \ \ \ \ 12x - 5y = 70$$$$2y = 11 - 3x$$$$y = \frac{11 - 3x}{2}$$$$12x - 5y = 70$$$$12x - 5\left(\frac{11 - 3x}{2}\right) = 70$$$$12x - \frac{55}{2} + \frac{15}{2}x = 70$$$$\frac{39}{2}x - \frac{55}{2} = 70$$$$\frac{39}{2}x = 70 + \frac{55}{2}$$$$\frac{39}{2}x = \frac{195}{2}$$$$x = 5$$$$y = \frac{11 - 3x}{2}$$$$y = \frac{11 - 3(5)}{2}$$$$y = -2$$Therefore, the coordinates of the center are,
$$C(\textcolor{#2192ff}{5}, \textcolor{#2192ff}{-2})$$Now let’s find the radius. Since $B$ lies on the circle, the distance from $C$ to $B$ is equal to the radius,
$$B (5, 11)\ \ \ \ \ \ C (5, -2)$$$$r = \sqrt{(5 - 5)^{2} + (-2 - 11)^{2}}$$$$r = \textcolor{red}{13}$$Therefore, the equation of the circle is,
$$(x - \textcolor{#2192ff}{5})^{2} + (y -\ (\textcolor{#2192ff}{-2}))^{2} = \textcolor{red}{13}^{2}$$$$(x - 5)^{2} + (y + 2)^{2} = 169$$Therefore, the final answer is,
$$(x - 5)^{2} + (y + 2)^{2} = 169$$3. A circle has center at the point $B(5, 1)$. The point $A(-1, -2)$ lies on the circle. Find the equation of the circle. (9709/12/O/N/20 number 9)

We already have the center, so we need to find the radius. The distance from the center, $B$, to the point $A$ is equal to the radius,
$$B(5, 1)\ \ \ \ \ \ A(-1, -2)$$$$r = \sqrt{(5 + 2)^{2} + (1 + 2)^{2}}$$$$r = 3\sqrt{5}$$Therefore, the equation of the circle is,
$$(x - 5)^{2} + (y - 1)^{2} = (3\sqrt{5})^{2}$$$$(x - 5)^{2} + (y - 1)^{2} = 45$$Therefore, the final answer is,
$$(x - 5)^{2} + (y - 1)^{2} = 45$$4. A circle with center $C$ has equation $(x - 8)^{2} + (y - 4)^{2} = 100$.

(a) Show that the point $T(-6, 6)$ is outside the circle. (9709/13/O/N/20 number 11)
$$(x - 8)^{2} + (y - 4)^{2} = 100$$Substitute the coordinates of point $T$ into the equation of the circle,
$$(\textcolor{#2192ff}{x} - 8)^{2} + (\textcolor{#0f0}{y} - 4)^{2}$$$$(\textcolor{#2192ff}{-6} - 8)^{2} + (\textcolor{#0f0}{6} - 4)^{2}$$$$200$$$$r^{2} = 200$$$$r = \sqrt{200}$$$$r = 14.1$$The radius of our circle is $\sqrt{100}$, which is $10$, since the line from the center to $T$ is larger than our radius $T$ lies outside the circle,
$$14.1 > 10$$Therefore, the final answer is,
$$14.1 > 10 \textmd{ hence }T(-6, 6)\textmd{ lies outside the circle}$$
Two tangents from $T$ to the circle are drawn.
(b) Show that the angle between one of the tangents and $CT$ is exactly $45^{\circ}$.

Let’s sketch a diagram of the problem,

Let’s start by finding the length of $CT$,
$$C(8, 4)\ \ \ \ \ \ T(-6, 6)$$$$CT = \sqrt{(8 + 6)^{2} + (4 - 6)^{2}}$$$$CT = 10\sqrt{2}$$Let’s use Pythagoras to find the angle $\theta$,
$$\cos{\theta} = \frac{10}{10\sqrt{2}}$$$$\theta = \cos^{-1}\left(\frac{10}{10\sqrt{2}}\right)$$$$\theta = 45^{\circ}$$Therefore, the final answer is,
$$\theta = 45^{\circ}$$The two tangents touch the circle at $A$ and $B$.
(c) Find the equation of the line $AB$, giving your answer in the form $y = mx + c$.
Let’s sketch a diagram of the problem,

Line $\textcolor{#0f0}{AB}$ is perpendicular to $\textcolor{red}{CT}$, so let’s find the gradient of $CT$,
$$C(8, 4)\ \ \ \ \ \ T(-6, 6)$$$$\textmd{gradient of CT } = \frac{6 - 4}{-6 - 8}$$$$\textmd{gradient of CT } = -\frac{1}{7}$$Since $AB$ is perpendicular to $CT$,
$$m_{1} \times m_{2} = -1$$$$-\frac{1}{7} \times m_{2} = -1$$$$m_{2} = 7$$Therefore, the gradient of $AB$ is,
$$7$$The midpoint of $CT$ lies on the line $AB$, so let’s find $M$,
$$M = \left(\frac{8 - 6}{2}, \frac{4 + 6}{2}\right)$$$$M = (1, 5)$$Therefore, the line $AB$ passes through,
$$M(1, 5)$$Now let’s find the equation of the line $AB$,
$$m = 7\ \ \ \ \ \ M(1, 5)$$$$y = mx + c$$$$5 = 7(1) + c$$$$5 = 7 + c$$$$c = 5 - 7$$$$c = -2$$$$y = 7x - 2$$Therefore, the final answer is,
$$y = 7x - 2$$(d) Find the $x$-coordinates of $A$ and $B$.

$A$ and $B$ are the points of intersection of the line $AB$ and the circle. Therefore, we will solve simultaneously,
$$y = 7x - 2\ \ \ \ \ \ (x - 8)^{2} + (y - 4)^{2} = 100$$Substitute $y = 7x - 2$ into the equation of the circle,
$$(x - 8)^{2} + (\textcolor{#2192ff}{y} - 4)^{2} = 100$$$$(x - 8)^{2} + (\textcolor{#2192ff}{7x - 2} - 4)^{2} = 100$$$$(x - 8)^{2} + (7x - 6)^{2} = 100$$Expand the quadratics,
$$x^{2} - 16x + 64 + 49x^{2} - 84x + 36 = 100$$Put all the terms on one side,
$$x^{2} + 49x^{2} - 16x - 84x + 64 + 36 - 100 = 0$$Solve the quadratic equation,
$$50x^{2} - 100x = 0$$$$50x(x - 2) = 0$$$$x = 0\ \ \ \ \ \ x = 2$$Therefore, the $x$-coordinates of $A$ and $B$ are,
$$x = 0\ \ \ \ \ \ x = 2$$