# 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

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

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 is *explicit – *y is the subject and is given in terms of x. The circle equation, on the other hand, is implicit – the $x$s and $y$s are mixed up.

The derivation of the equation of a circle is from an application of Pythagoras Theorem. Specifically, the longest side squared is equal to the sum of the squares of the shorter sides. Evidently, the longest side is the radius. If we pick a point on the circle $(x,y)$, we see that the shorter sides have lengths $x-a$ and $y-b$. It follows that $(x-a)^2+(y-b)^2=r^2$.

They may present you 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.

### 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

- 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.
- Find the equation of the circle given that the centre is at (1,2) and the point (3,5) lies on the circle.
- Find the equation of the tangent to the circle $(x-4)^2+(y-1)^2=10$ at the point (3,4).

Draw your own sketches to help you with these questions.

- $(x-2)^2+(y+5)^2=11$.
- $(x-1)^2+(y-2)^2=13$ 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 $y=-\frac{1}{3}x+3$.

### (Completing the Square)

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.

Answer:

- 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 $x$.

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 $-4/(x+2)$. Using rise over run 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 of x is $x=\frac{3-\sqrt{33}}{2}$.

### (Perpendicular Bisector)

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

Always begin by sketching the information given. We can find the point of intersection of the chord with the diameter since, if the intersection is at right angles, then they will intersect at the midpoint of the chord. It follows that 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 is $(-2,-2)$ 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 $x=1+\sqrt{7}$.

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