Write the equation in standard form for the circle x2=–y2+12y+28.

The given equation is x^2 = -y^2 + 12y + 28.

First, we need to rewrite the equation in the form of a standard equation of a circle, which is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

To do this, we need to complete the square for the y terms.

The equation can be rewritten as:

x^2 = -(y^2 - 12y) + 28

To complete the square, we take half of the coefficient of y, square it and add it to both sides. Half of -12 is -6, and (-6)^2 = 36.

x^2 + 36 = -(y^2 - 12y + 36)

This simplifies to:

x^2 + 36 = -(y - 6)^2

We want to get rid of the negative sign in front of the y term, so we multiply the entire equation by -1:

-x^2 - 36 = (y - 6)^2

Finally, we rearrange the equation to match the standard form:

(x - 0)^2 + (y - 6)^2 = 36

So, the standard form of the given equation is (x - 0)^2 + (y - 6)^2 = 36, which represents a circle with center at (0,6) and radius 6.