Question

How do you write the equation `4x^2+4y^2-24y+36=0` in standard form and find the center and radius?

Answer 1

Given `4x^2+4y^2-24y+36=0" [1]"`

The a conic section of the general form:

`Ax^2+Bxy+Cy^2+Dx + Ey + F = 0" [2]"`

Matching coefficients between equations [1] and [2] we observe that:

A = B = 4#

This identifies the given conic section as a circle of the general form:

`(x-h)^2+(y-k)^2 = r^2" [3]"`

Begin progress to the form of equation [3] by dividing equation [1] by 4:

`x^2+y^2-6y+9=0" [1.1]"`

Subtract 9 from both sides:

`x^2+y^2-6y= -9" [1.2]"`

Because there is no x term, we conclude that `h = 0`:

`(x-0)^2+(y-k)^2 = r^2" [3.1]"`

There is a y term, therefore, we add k^2 to both sides of equation [1.2]:

`x^2+y^2-6y + k^2= k^2-9" [1.3]"`

From the pattern `(y - k)^2 = y^2- 2ky + k^2`, we know that we can can find the value of k by setting middle term of the parttern equal to the y term:

`-2ky = -6y`

`k = 3 larr` substitute this into equations [3.1] and [1.3]:

`(x-0)^2+(y-3)^2 = r^2" [3.2]"`

`x^2+y^2-6y + 3^2= 3^2-9`

Simplify:

`x^2+y^2-6y + 9 = 0 [1.4]"`

Because the right side of equation [1.4] this indicates that the radius is 0:

`(x-0)^2+(y-3)^2 = 0" [3.3]"`

This degenerates into the point `(0,3)`