Circle has the equation #x^2+y^2+2x-2y-14=0#, how do you graph the circle using the center (h,k) radius r?

1 Answer
Dec 5, 2015

Answer:

I found a circle centred at #(-1,1)# and with radius #r=4#.

Explanation:

I would write:
#(x^2+2x)+(y^2-2y)=14#
add and subtract some values:
#(x^2+2xcolor(red)(+1-1))+(y^2-2ycolor(red)(+1-1))=14#
#(x^2+2x+1)+(y^2-2x+1)-2=14#
#(x+1)^2+(y-1)^2=16#
or
#(x-(-1))^2+(y-(1))^2=16#
#(x-(-1))^2+(y-(1))^2=4^2#
Which is in the form:
#(x-h)^2+(y-k)^2=r^2#
giving you a circle centred at #(-1,1)# and with radius #r=4#.

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