Equation Of A Circle | Studywell.com

Equation of a Circle

Equation of a Circle Notes

We find the equation of a circle from the coordinates of its centre and its radius. Consider this circle whose centre is at the point (a,b) and whose radius is r. The equation of the circle is given by

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

For a particular circle, the values of $a$ , $b$ and $r$ are specified whereas $x$ and $y$ are left as general points. This is much like when we specify $m$ and $c$ in $y=mx+c$ when we have the equation of a given straight line. Note that the equation of a straight line in this form is explicit – this means that y is the subject and is given in terms of x. The circle equation, on the other hand, is implicit – put simply the x and y appear anywhere in the equation.

The derivation of the equation of a circle is from an application of Pythagoras and can be seen by drawing a right-angled triangle such that the radius is the hypoteneuse – see the derivation here. If we pick any point on the circle $(x,y)$ , we see that the shorter sides have lengths $x-a$ and $y-b$ while the hypoteneuse has length $r$ . Since this is true for any point on the circle it follows that $(x-a)^2+(y-b)^2=r^2$ .

You will be presented with questions that expect you to know the equation of a circle. Other questions might bring in knowledge from other areas of maths such as finding mid-points. In more complicated questions they may ask you to find gradients using the knowledge that a tangent to a circle is perpendicular to its radius. Before we try some examples, we will look at some of the other things you need to know about circles.

Things to Learn for Circle Equation Questions

You should know the following facts:

Most importantly, the equation of a circle.
Secondly, that the angle subtended from a diameter at the circumference is a right angle.
Next, the radius and tangent touch at right angles to one another.
Finally, a perpendicular from a chord bisects the chord.

You should be able to find the equation of a circle:

using completing the square and/or Pythagoras,
given a triangle that has all three points on its circumference and
find the equation of a tangent using perpendicular gradients.

Examples

1. The centre of a circle is given by (2,-5) and its radius is the square root of 11. Write down the equation of the circle.
2. Find the equation of the circle given that the centre is at (1,2) and the point (3,5) lies on the circle.
3. Find the equation of the tangent to the circle $(x-4)^2+(y-1)^2=10$ at the point (3,4).
4. The equation of a circle is given by $x^2+6x+y^2-4x+9=0$ . Find the centre and the radius of the circle. Find all points on the circle with $x$ -coordinate $-4$ .

Draw your own sketches to help you with these questions.

Solution:

$(x-2)^2+(y+5)^2=11$ .
$(x-1)^2+(y-2)^2=13$ using Pythagoras to find the radius.
Consider the radius joining the centre at (4,1) with the point (3,4). The gradient of this radius is -3 and so the tangent will have gradient $\frac{1}{3}$ . The equation of the tangent so far is $y=\frac{1}{3}x+c$ and we find $c$ by inserting the coordinates (3,4) giving $c=3$ . The equation of the tangent is $y=\frac{1}{3}x+3$ .
Rearrange the equation by completing the square: $(x+3)^2-9+(y-2)^2-4+9=0$ or $(x+3)^2+(y-2)^2=4.$ Hence, the centre is at (-3,2) and the radius is 2. Substituting $x=-4$ into this equation gives $1+(y-2)^2=4$ . It follows that the y coordinates are given by $2\pm\sqrt{3}$ and so the coordinates are $(-4,2+\sqrt{3})$ and $(-4,2-\sqrt{3})$ .

The three points $(1,-2)$ , $(4,5)$ and $(5,x)$ lie on the circumference of a circle. The points $(1,-2)$ and $(4,5)$ are at either end of the diameter. Find the exact value of $x$ .

Solution:

Always sketch when answering coordinate geometry questions. Since the line joining $(1,-2)$ and $(4,5)$ is a diameter, the line joining $(1,-2)$ and $(5,x)$ is at right angles to the line joining $(4,5)$ and $(5,x)$ . The gradient of the line joining $(1,-2)$ and $(5,x)$ is $(x+2)/4$ (using rise over run). Hence, the line joining $(4,5)$ and $(5,x)$ is perpendicular and has gradient $-4/(x+2)$ . Using rise over run again to get another expression for this gradient we have $\frac{x-5}{1}=-\frac{4}{x+2}$ or $(x-5)(x+2)=-4$ . It follows that solving the quadratic $x^2-3x-6=0$ gives $x=\frac{3\pm\sqrt{33}}{2}$ . The y-coordinate is negative and so the exact value is $x=\frac{3-\sqrt{33}}{2}$ .

The diameter of the circle whose centre is at $\left(1,-\frac{5}{2}\right)$ is perpendicular to the chord that joins the circumferential points $(x,-4)$ and $(-2,-2)$ . Find the value of $x$ .

Solution:

Always begin by sketching the information given. We can find the point of intersection of the chord with the diameter since, if the lines are perpendicular, then the diameter will intersect at the midpoint of the chord. The midpoint can be expressed as $\left(\frac{x-2}{2},-3\right)$ . The line between this midpoint and the centre is perpendicular to the line between the midpoint and $(-2,-2)$ , for instance. Firstly, the gradient of the line between the midpoint and the centre is $\frac{-\frac{5}{2}+3}{1-\frac{x-2}{2}}=\frac{0.5}{2-\frac{x}{2}}=\frac{1}{4-x}$ . Secondly, the gradient between the midpoint and the point $(-2,-2)$ is $\frac{-3+2}{\frac{x-2}{2}+2}=\frac{-2}{x+2}$ . Note that the product of perpendicular gradients is -1 – this can be seen from $m\times -\frac{1}{m}$ . It follows that $\frac{1}{4-x}\times \frac{-2}{x+2}=-1$ or $(4-x)(x+2)=2$ . Solving this quadratic gives $x=1\pm\sqrt{7}$ . From our sketch we can see that $x$ is positive and so $x=1+\sqrt{7}$ .

Videos

https://youtu.be/pdEgKNc7RZs

Past exam question and solutions on the equation of a circle including the equation of a tangent.

https://youtu.be/9HX7VkBZE3k

Past AS maths exam question on circle geometry which is also heavy on trigonometry.

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