How do you find the center and radius for #x^2 + y^2 +2x -2y -14 = 0#?

1 Answer
Aug 6, 2016

Answer:

The circle has the center in #(-1;1)# and the radius of #4#

Explanation:

To find the center and radius we have to transform the equation to form:

#(x-a)^2+(y-b)^2=r^2 #

We start from:

#x^2+2x+y^2-2y-14=0#

We transform it to:

#x^2+2x color(red)(+1)+y^2-2ycolor(red)(+1)-14 color(red)(-2)=0#

I added #1# to transform the formulas to #(a+-b)^2#. If I added #2# then I had to substract #2# to keep the equation balanced. These operation are marked in red.

#(x+1)^2+(y-1)^2-16=0#

#(x+1)^2+(y-1)^2=16#

From this equation we read that: the centre is #(-1;1)# and the radius is #r=4#