Illustrating formation of equation of a circle

The equation of a circle

The equation of a circle is a fundamental concept especially in coordinate geometry. The equation represents all the points that are equidistant from a fixed point known as the center of a circle in a two-dimensional plane.

Consider a circle of radius r on a Cartesian plane centered at point o(a,b) as shown in figure below. A point p on the circumference of the circle has an arbitrary point (x, y).

line OQ an QP makes a right angle triangle with the radius r of the circle such OP=r, OQ = x-a and QP=y-b.

using Pythagoras’ theorem: (x-a)2 + (y-b)2 = r2.

Hence the general equation of a circle is given as:

(x-a)2 + (y-b)2 = r2.

where:

  • r is the radius of the circle
  • (a, b) are the coordinates at the center of the circle
  • (x, y) is an arbitrary point on the circumference of the circle.

The equation of a circle problems

Find the equation of a circle centered at (4, 5) and with radius of 3 units.

Solution

Recall the equation: (x-a)2 + (y-b)2 = r2

a=4, b=5, r=3

hence (x-4)2+(y-5)= 32

=x2– 4x- 4x+ 16+ y2– 5y- 5y + 25= 9

x2-8x + 16 + y2 – 10y + 25 = 9

x2– 8x + y2– 10y+ 41=9

x2 – 8x + y2 – 10y + 32=0

Example problem

Find the equation of a circle given the center of the circle in the Cartesian plane and radius in each of the following:

(i) center : (0,2), radius = 3 units

(ii) center: (-5, -3), radius = 6 units

$$ (iii) center: \ (3, 2) , radius = \frac{sqrt{7}}{3} $$

solutions

(i) taking an arbitrary point on the Cartesian plane: (x, y); we have:

$$ (x – 0)^2 + (y-2)^2 = 3^2$$
$$x^2 +y(y-2)-2(y-2) =9$$ $$=x^2+y^2-2y-2y+4=9$$ $$=x^2+y^2-4y+4-9=0$$

hence equation of our circle on the Cartesian plane will be: x2 +y2-4y-5=0


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