Determine the equation of the circle expressing it in the form x^2+y^2+ax+by+c=0 where a,b and c are integral values

Answer 1

To determine the equation of a circle in the form x^2 + y^2 + ax + by + c = 0, we need to identify the values of a, b, and c.

The general equation of a circle is:
(x - h)^2 + (y - k)^2 = r^2

In this form, (h, k) represents the center of the circle and r represents the radius.

First, we expand the equation:
x^2 + y^2 + ax + by + c = 0

Next, we complete the square for x and y terms:
x^2 + ax + y^2 + by = -c

To complete the square for x terms, we take half the coefficient of x, square it, and add it to both sides:
(x^2 + ax + (a/2)^2) + y^2 + by = -c + (a/2)^2

Simplifying the x terms, we get:
(x + a/2)^2 + y^2 + by = -c + (a/2)^2

Now, we complete the square for y terms:
(y^2 + by + (b/2)^2) + (x + a/2)^2 = -c + (a/2)^2 + (b/2)^2

Simplifying the y terms, we get:
(y + b/2)^2 + (x + a/2)^2 = -c + (a/2)^2 + (b/2)^2

Comparing this with the general equation of a circle, we can determine the values of a, b, c:
h = -a/2, k = -b/2, r^2 = -c + (a/2)^2 + (b/2)^2

Therefore, the equation of the circle in the form x^2 + y^2 + ax + by + c = 0 can be rewritten as:
(x + a/2)^2 + (y + b/2)^2 = -c + (a/2)^2 + (b/2)^2