Geometry of circles
Posted in Maths, Core 1The equation of a circle is given as:
(x - a)2 + (y - b)2 = r2
Where the centre of the circle is (a, b) and the radius is r. Hence a circle with a centre of the origin (0, 0) has an equation of:
x2 + y2 = r2
There is an alternative equation of a circle:
x2 + y2 + 2gx + 2fy + c = 0
This is known as the general form and is simply an expansion of the centre and radius form of the equation. To turn an equation in the general form into the radius and centre form you need to complete the square:
Example question: Write 4x2 + 4y2 - 16x + 8y - 80 = 0 in the form (x - a) + (x - b) = r2 and hence write the centre and radius of the circle.
Worked answer: Firstly you should make the coefficients of both x2 and y2 equal to 1, in this example you divide by 4 to leave you with:
x2 + y2 - 4x + 2y - 20 = 0
Then you should rewrite the equation by grouping the x and y terms together:
x2 - 4x + y2 + 2y - 20 = 0
Now you complete the square, twice, for both the x terms and y terms:
(x - 2)2 + (y + 1)2 = 0
To finish completing the square we need to add a constant back. From the first bracket the constant created when expanded is +4 and for the second bracket is +1 giving a total of +5 from the brackets. The original constant was -20 so we must subtract 25:
(x - 2)2 + (y + 1)2 - 25 = 0
Then simply add +25 to both sides:
(x - 2)2 + (y + 1)2 = 25
The equation is now in the correct form so we can easily find the centre as (2, -1) and the radius as 5.
Normals to circles
The normal at a given point on a circle is the line which cuts through the circle at right angles:
The equation of the circle is found using the 'equation of a line between two points' formula from the geometry of straight lines topic. Ignore the circle and what you are left with is a straight line passing through two points, point P(x1, y2) and point C(x2, y2), the centre of the circle. The equation of the normal is therefore:
$[page]\fs5\frac{y - y_1}{y_2 - y_1}\text{ = }\frac{x - x_1}{x_2 - x_1}[/page]$Tangents to circles
The tangent at a given point P on a circle is the point where a line just touches the edge of the circle:
The important thing to realise about tangents is that they cross the normal and meet the radius at the same point, at right angles. As the normal and tangent are perpendicular their gradients multiply to give -1 and this helps to calculate the equation of a tangent to a circle:
Worked example
Example question: Find the equation of the tangent to the circle (x - 2)2 + (y - 1)2 = 25 at point (6, 4)
Worked answer: Firstly you need to find the centre of the circle, from the equation of the circle this is (2, 1).
Now you need to find the gradient of the normal. The gradient is the difference in y co-ordinates divided by the difference in x co-ordinates and the normal passes through both the centre (2, 1) and the point (6, 4). Therefore the gradient is:
As previously mentioned the gradients of two lines which are perpendicular to each other multiply to give -1. Therefore the gradient of the tangent which crosses at right angles to the normal is or
. The tangent passes through the point (6, 4). The equation of a line with gradient m through a point (x1, y2) is y - y1 = m(x - x1). Substituting in our numbers: