Write the equation of the circle in standard form. x^2+y^2-6x+8y+9=0.

We'll recall the standard form of the circle
equation:

(x - h)^2 + (y - k)^2 =
r^2

h and k represent the coordinates of the center of the
circle and r is the value of the radius of the circle.

To
reach to this form, we'll have to complete the
squares:

(x^2 - 6x + ...) + (y^2 + 8y + ...) =
-9

We'll consider the
formula:

(a+b)^2 = a^2 + 2ab +
b^2

2xb = -6x

b = -3 =>
b^2 = 9

x^2 - 6x + ... = x^2 - 6x +
9

y^2 + 8y + ... = y^2 + 8y +
16

(x^2 - 6x + 9) + (y^2 + 8y + 16)  -9 -16 =
-9

We'll move the numbers to the right
side:

(x^2 - 6x + 9) + (y^2 + 8y + 16) = 9 + 16 -
9

(x^2 - 6x + 9) + (y^2 + 8y + 16) =
16

(x-3)^2 + (y + 4)^2 =
4^2

The standard form of the equation of the
circle is: (x-3)^2 + (y + 4)^2 =
4^2