Find the equation of a circle and its center and radius if the circle passes through the points (3 , 2) , (6 , 3) and (0 , 3).

The general equation of a circle is given by the equation: Ax^{2} + Ay^{2} + Bx + Cy + D = 0

Because each point given should fulfill the equation of the circle we have to solve the following set of equations with the unknowns A, B, C and D:

A(x^{2}+ y^{2}) + Bx + Cy + D = 0

A(x_{1}^{2}+ y_{1}^{2}) + Bx_{1}+ Cy_{1}+ D = 0

A(x_{2}^{2}+ y_{2}^{2}) + Bx_{2}+ Cy_{2}+ D = 0

A(x_{3}^{2}+ y_{3}^{2}) + Bx_{3}+ Cy_{3}+ D = 0

or

Because all the equations equal to 0 also the determinant of the coefficients should be equal to 0 and the value of the determinant will be as follows:

Notice that we got the equation of a circle with the determinants equal to the coefficients A, B, C and D as follows; (x^{2}+ y^{2})A + xB + yC + D = 0.

After dividing all terms by 6 we get: A = 1 B =−6 C = −14 D = 33.

And the equation of the circle is: x^{2}+ y^{2}⎯ 6x ⎯ 14y + 33 = 0

In order to find the radius of the circle use the general circle equation and perform some basic algebraic steps and with the help of the square form (a + b)^{2} = a^{2} + 2ab + b^{2} we get

The last equation is a circle with the center and radius equals to (notice the minus sign at x and y):

and

and the radius is:

The equation of the circle can be presented by the center and the radius as: (x ⎯ 3)^{2} + (y ⎯ 7)^{2}= 5^{2}