Circle's equation.Find the equation of the circle that passes through (8, 0), (0, 6) and (0, 0).

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The standard equation of a circle is x^2 + ax + y^2 + bx + c = 0

As it passes through (8, 0), (0, 6) and (0, 0) substituting the coordinates of the points in the equation we get 3 equations

c = 0 ...(1)

64 + 8a  = 0 ...(2)

36 + 6b = 0 ...(3)

(2)

=>a = -64/8 = -8

(3)

=> b = -36/6 = -6

The equation of the circle is x^2 + y^2 -8x -6y = 0

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll use the determinant:

 

x^2 + y^2    x    y    1

det A = 8^2 + 0^2    8    0    1 = 0

0^2 + 6^2    0    6    1

0^2 + 0^2    0    0    1

 

x^2 + y^2    x    y    1

det A =       64           8    0    1  = 0

36           0    6    1

0            0    0    1

We'll compute the determinant considering the last row, since it has a lot of zeroes.

det A = (-1)^(4 + 4)*1*det A1

x^2 + y^2    x    y

det A1 =           64         8    0

36         0    6

det A1 = 48(x^2 + y^2) - 288y - 384x

The equation of the circle that passes through the given points is:

48x^2 + 48y^2 - 288y - 384x = 0

We’ve answered 318,916 questions. We can answer yours, too.

Ask a question