How do you find the equation of a circle in standard form given #x^2+y^2+4x+6y+4=0#?

The standard Cartesian form of the equation of a circle is:

#(x-h)^2 + (y-k)^2 = r^2" [1]"#

Expand the squares:

#x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"#

The given equation is:

#x^2 + y^2 + 4x + 6y + 4 = 0" [3]"#

To make equation [3] look more like equation [2] group the x term together, group the y terms together and move the constant term to the right side:

#x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"#
#x^2 + 4x + y^2 + 6y = -4" [4]"#

Add #h^2 and k^2# to both sides of equation [4]

#x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"#
#x^2 + 4x + h^2 + y^2 + 6y + k^2 = h^2 + k^2 -4" [5]"#

To find the value of "h", set the second term of equation [2] equal to the second term of equation [5] and then solve for h:

#-2hx = 4x#

#h = -2#

To find the value of "k", set the fifth term of equation [2] equal to the fifth term of equation [5] and then solve for k:

#-2ky = 6y#

#k = -3#

Now that we know the values of h and k, we know that we can substitute #(x - -2)^2# for the first 3 terms and #(y - -3)^2# for the next 3 terms into equation [5]:

#(x - -2)^2 + (y - -3)^2 = h^2 + k^2 - 4" [6]"#

Substitute -2 for h and -3 for k into equation [6]:

#(x - -2)^2 + (y - -3)^2 = (-2)^2 + (-3)^2 - 4" [7]"#

#(x - -2)^2 + (y - -3)^2 = 4 + 9 - 4" [8]"#

Combine the constants and write the constant as a square:

#(x - -2)^2 + (y - -3)^2 = 3^2" [9]"#

Equation [9] is in standard Cartesian form.