How do you find the standard form given #x^2+y^2-8x+4y+11=0#?

1 Answer
Aug 2, 2017

Answer:

#(x-4)^2 + (y+2)^2 = 9#

Explanation:

Firstly, which conic is this?

Well, the coefficients in front of the #x^2# and #y^2# terms are the same, and there is a plus sign between them, so it is a circle.

The standard form equation of a circle is given by

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

where

  • #h# is the #x#-coordinate of the center of the circle

  • #k# is the #y#-coordinate of the center of the circle

  • #r# is the radius of the circle

We'll need to complete the square to find the standard form equation.

Let's move the constant #11# to the right hand side, and group #x#- and #y#-terms together:

#x^2 - 8x + y^2 + 4y = -11#

Set up the completion of the square for #x# and #y#:

#(x^2 - 8x + ul(" ")) + (y^2 + 4y + ul(" ")) = -11#

We're looking for the "#c#" term in the parabolic equation, and to find that we do

#ul(c = (b/2)^2#

or

#ul(c = (b^2)/4#

The #b# term for the #x# is #-8# (the variable in front of the linear term #x#), so we have

#x{color(white)(a)c = ((-8)^2)/4 = color(red)(ul(16#

Similarly, for the #y#, #b = 4#, so we have

#y{color(white)(a)c = (4^2)/4 = color(green)(ul(4#

Now, we plug these values into the equation:

#(x^2 - 8x + color(red)(16)) + (y^2 + 4y + color(green)(4)) = -11#

Now, we add both values of #c# to the constant on the right side:

#(x^2 - 8x + color(red)(16)) + (y^2 + 4y + color(green)(4)) = -11 + color(red)(16) + color(green)(4)#

Each of the terms can now be represented as the square of a monomial, and so we have

#color(blue)(ulbar(|stackrel(" ")(" "(x-4)^2 + (y+2)^2 = 9" ")|)#

(to get the value of #h# and #k#, simply take half of the #b# value for each variable; i.e. #overbrace(-8/2 = -4)^x# and #overbrace(4/2 = 2)^y#

For this equation here, we can determine:

  • the center point #(h,k)# of the circle is #(4, -2)#

  • the radius #r# is #sqrt9 = 3#