How do you find the standard form given #x^2+y^28x+4y+11=0#?
1 Answer
Answer:
Explanation:
Firstly, which conic is this?
Well, the coefficients in front of the
The standard form equation of a circle is given by
#ul((xh)^2 + (yk)^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
#x^2  8x + y^2 + 4y = 11#
Set up the completion of the square for
#(x^2  8x + ul(" ")) + (y^2 + 4y + ul(" ")) = 11#
We're looking for the "
#ul(c = (b/2)^2#
or
#ul(c = (b^2)/4#
The
#x{color(white)(a)c = ((8)^2)/4 = color(red)(ul(16#
Similarly, for the
#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
#(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(" ")(" "(x4)^2 + (y+2)^2 = 9" "))#
(to get the value of
For this equation here, we can determine:

the center point
#(h,k)# of the circle is#(4, 2)# 
the radius
#r# is#sqrt9 = 3#