For the circle x^2+y^2+2x-6y-6=0 of the previous problem, find the center.?

Answer 1

the standard equation af the circle is
(X-a)^2+(Y-b)^2= c^2
where a,b and c are constans
ant the radius = c
and the center is at point (a,b)
so all what u have to do is to put the given equation on that shape and get what u want

THAT WOULD BE
X^2+2X+Y^2-6Y-6=0
(X+1)^2 - 1 + (Y-3)^2 -9 -6 =0
SO ...,
(X+1)^2+(Y-3)^2=16
(X+1)^2+(Y-3)^2 =4^2
THEN
compare with the equation
(X-a)^2+(Y-b)^2= c^2
the radius of that circle is c = 4
and the center is point (a,b)that is (-1,3)


NOTE THAT
the term (X^2+aX) = (X+(a/2)X)^2 - a^2

Answer 2

Dude! if you paid attention in math class this would be no problem. Okay this how it goes.....

x^2+2x+y^2-6y=6 Then, i put like terms together.
x^2+2x/2+y^2-6y/2=6+1^2+3^2 Over here, notice how i divided by 2 for each coefficient with a variable and the answer i squared it on the other side of equation. For ex: 2x/2=1 there fore on the other side of equation i put 1^2.
(x+1)^2+(y-3)^2=16 notice that the standard form of the equation of circle has been found. (x-h)^2+(y-k)^2=r^2
Thus, x=-1 and y=3 as the center of circle.

Now the square root of the radius which will be =4, this indicates the number of spaces you must count after plotting the center point on the graph.

I hope this helped. GO TO CLASS!!!!

Answer 3

The above equation can be written as: x^2+2x+1+y^2-6y+9-16=0
Now this can be written as: (x+1)^2+(y-3)^2=16
Hence the center is (-1,3) with radius 4