Equation of a Circle

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What is the equation of a circle?

A circle with centre (a, b) and radius r has the equation

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

Circle with centre (a,b) and radius r

You need to be able to find the equation of a circle given its centre and radius
- Substitute the values into the formula

Finding the equation of a circle

How do I find the centre and radius of a centre given its equation?

Make sure it is in the form ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$
- The radius is the positive square root of the constant term
- The coordinates of the centre can be found by finding the values that make each bracket equal to zero

Finding the centre and radius of a circle given its equation

Examiner Tip

Remember that the numbers in the brackets have the opposite signs to the coordinates of the centre
Don't forget to take the square root of the right-hand side of the equation when finding the radius

Worked example

Eqn of Circle Example, A Level & AS Level Pure Maths Revision Notes

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Finding the Centre & Radius

What are the different forms of the equation of a circle?

The most useful equation of a circle is ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$
- This is so the centre, $(a, b)$ and radius $r$ are easy to see
Any other form of the equation of a circle can be rearranged into this form
- The most common alternative form for the equation of a circle is called the general form
  $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

How do I find the centre and radius of a circle from any form of its equation?

A circle equation in a different form can always be rearranged into (x- a)² + (y - b)² = r²
- The centre is then $(a, b)$ and radius $r$
Rearranging to this form will often involve completing the square

Completing the square to find the centre and radius of a circle

Worked example

Circle Ctr Rad Example, A Level & AS Level Pure Maths Revision Notes

Intersection of a Circle & a Line

What is meant by the intersection of a circle and a line?

A line may pass through a circle
- in which case it will intersect the circle twice
- the part of the line between the two points of intersection will be a chord
  - or, if it passes through the centre of the circle, a diameter
A line may touch a circle
- in which case it will intersect the circle once
- such a line would be called a tangent to the circle
A line may not intersect a circle at all

The three cases for intersections between a circle and a lin

How do I determine whether a line and a circle intersect?

For the equation of a circle in the form ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ and the equation of a line in the form $y = m x + c$
STEP 1
Substitute the linear equation into the circle equation
- e.g.
  ${(x - 5)}^{2} + {(y - 2)}^{2} = 13$ and $y = x - 4$
  would become
  ${(x - 5)}^{2} + {((x - 4) - 2)}^{2} = 13$
STEP 2
Expand, rearrange and simplify this equation - it should be a quadratic
- e.g.
  ${(x - 5)}^{2} + {(x - 6)}^{2} = 13$
  $x^{2} - 10 x + 25 + x^{2} - 12 x + 36 - 13 = 0$
  $x^{2} - 11 x + 24 = 0$
STEP 3
Solve the equation to deduce the number of intersections
If there are two solutions, there are two intersections, one solution (repeated) indicates a tangent, no (real) solutions indicates no intersection
- e.g.
  $(x - 3) (x - 8) = 0 x = 3, x = 8$
  Two solutions so the line and the circle intersect twice
STEP 4
If required, find the $y$ -coordinates of the intersection(s)
- e.g.
  $y = x - 4 y = 3 - 4 = - 1, y = 8 - 4 = 4$
  The line and the circle intersect at the points $(3, - 1)$ and $(8, 4)$

Examiner Tip

A horizontal or vertical line could intersect a circle, when the full method shown above is unnecessary
- Horizontal lines have the form $y = k$ , so substitute $k$ for $y$ in the circle equation (and solve for $x$ )
- Vertical lines have the form $x = k$ , so substitute $k$ for $x$ in the circle equation (and solve for $y$ )

Worked example

Show that the line $y = 4 x + 4$ is tangent to the circle ${(x - 4)}^{2} + {(y - 3)}^{2} = 17$ .
State the coordinates of the point of intersection between the tangent and the circle.

STEP 1 - Substitute the linear equation into the circle equation

${(x - 4)}^{2} + {(4 x + 4 - 3)}^{2} = 17$

STEP 2 - Rearrange to a quadratic

$\begin{array}{rcl} {(x - 4)}^{2} + {(4 x + 1)}^{2} - 17 & = & 0 \\ x^{2} - 8 x + 16 + 16 x^{2} + 8 x + 1 - 17 & = & 0 \\ 17 x^{2} & = & 0 \end{array}$

STEP 3 - Solve

$x^{2} = 0$

$x = 0$ (repeated)

There is only one point of intersection so the line $y = 4 x + 4$ is tangent to the circle ${(x - 4)}^{2} + {(y - 3)}^{2} = 17$

STEP 4 - The coordinates are required

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

The line and circle intersect at $(0, 4)$

Equation of a Circle (Cambridge O Level Additional Maths)

Revision Note