Geometry of the Line (4): Circles

Time to go back to straight line geometry for a bit.

With knowledge of how to calculate the gradient, the perpendicular gradient, the mid-point and the length of the line, you've also done quite a lot of co-ordinate geometry of the circle. This is an interesting concept. A circle on a set of cartesian co-ordinates.

You start from the premise of a circle with centre (0,0). The equation of this circle is x² + y² = r². If you study a diagram of it, or even draw one yourself, the reason for this becomes clear. Like a lot of co-ordinate geometry, it's based on Pythagoras. In this case, drawing a triangle inside the circle to a point somewhere on the circumference. Say that this circle with centre (0,0) passes through the point (3,4).

The line from the centre to the point functions as the hypoteneuse of the right angled triangle, while line x measures the distance travelled horizontally to the point, and the line y measures the vertical distance to the point.

The equation of this particular circle then tells us the radius.

3² + 4² = r²

or 9 + 16 = r²

= 25

r = 5.

The radius of the circle is 5.

Things get a little more complicated if the circle's centre is not (0,0). Say we have a circle with centre (2,3). We still devise the equation of the circle based on its distance from (0,0).

This circle's centre has moved 2 units x-wards and 3 y-wards (across and up) from the origin. Any calculation of the radius needs to take into account the fact that we are already some way from the origin.

So if the circle passes through the point (5,8), the distance from the centre to the point will be 5-2 and 8-3 (otherwise we're talking about including space outside the circle).

So the equation will be: (x-2)² + (x-3)² = r².

In this case: 9 + 25 = 34

So r = √34

It is perfectly acceptable to leave the answer as a surd at C1, indeed often you have to.

Usually you are given the equation, say (x-4)² + (y-7)² = 4

or something.

The centre of the circle is ALWAYS THE OPPOSITE OF THE NUMBERS IN BRACKETS, in this case (4,7). Worth remembering. If it is (x-a)² + (y-b)² = r² the centre is always (a,b).