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

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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 **

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**