Question

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

Answer 1

Please see the explanation.

Explanation:

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.