The equation of the circle is related to the Pythagoras' Theorem.

①At any point (x,y) on the circumference, you can always draw a line from the circle's centre to it. No matter which point you pick, this line is always a radius of the circle.

②you can always form a right-anglead triangle with the following three :

(a)The radius
(b)Either the x or y - axis
(c)A horizontal or vertical line

(a) is the hypotenuse of the triangle.

The 3 vertices of the triangle are the centre, a point on the y/xaxis, and the point on the circumference.


By Pythagoras' Theorem,
a + b² = c²

For this triangle,

a is the horizontal distance between the centre and the point on the circumference.

b is the vertical distance between the centre and the circumference.

c is the radius (length of the hypotenuse)

So,

For any point (x,y) on the circumference of a circle with a centre (a,b) and radius r,


The horizontal distance between it and the centre = (x - a) units


The vertical distance between it and the centre = (y - b) units

(Now you may get a negative value for them depending on the point chosen. but it doesn't matter as we are only interested in its absolute value.)

We can form a right angle triangle with sides (x - a) units, (y - b) units and radius r units.


Using Pythagoras' Theorem,

(x - a)² + (y - b)² = r²

Notice how the negative values are now positive since they have been squared.


The general form is merely the expanded form.


x² - 2ax + a² + y² - 2by + b² = r²

x² + y² - 2ax - 2by + a² + b² - r² = 0

Rewrite as
x² + y² + (-2a)x + (-2b)y + (a² + b² - r²) = 0

if you compare this to the one given in the textbook,

x² + y² + 2gx + 2fy + c = 0

Comparing coefficients,

-2a = 2g →a = -g

-2b = 2f → b = -f

So the coordinates of the centre is simply (-g,-f)


c = a² + b² - r²

r² = a² + b² - c

r = √(a² + b² - c)

The whole idea of the question is to rearrange and rewrite it in another way such that it is comparable to the two given forms.

Then you compare the coefficients to find your radius and centre.