1.3.2 Equation of a Circle - GCEALevelMaths9709

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,

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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,

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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,
$(\textcolor{#2192ff}{x} - 8)^{2} + (\textcolor{#0f0}{y} - 4)^{2}$ $(\textcolor{#2192ff}{-6} - 8)^{2} + (\textcolor{#0f0}{6} - 4)^{2}$ $104$ $r^{2} = 104$ $r = 2\sqrt{26}$ $r = 10.2$ 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,
$10.2 > 10$ Therefore, the final answer is,
$10.2 > 10 \textmd{ hence }T(-6, 6)\textmd{ lies outside the circle}$
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,

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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,

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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,
$x = 0\ \ \ \ \ \ x = 2$

GCE_AS_Level_Coordinate_Geometry__Equation_of_a_Circle Download