Question

an equation of a circle is x2 + y2 + 10x – 6y + 18 = 0.


A.Show all your work in determining the center and radius of this circle.

B.In complete sentences, explain the procedure used.

Answer 1

general eq of circle
x2+y2+2gx+2fy+c=0
center of circle is C=(-g,-f)
radius of circle is r = (g2+f2+c)^1/2
comparing this with above equation
x2+y2+2(5)x+2(-3)+18=0 g=5 f=-3 C=(-5,3)
for radius
g=5 f= -3 c=18 r =(25+9+324)^1/2 r=(358)^1/2

Answer 2

I am assuming you mean x^2 + y^2 + 10x + 6y + 18 =0 is the equation:
THEN;
x^2 + y^2 + 10x + 6y + 18 = 0
x^2 + 10x + y^2 + 6y + 18 = 0 [just reaarranging]
x^2 + 10x +25 + 6y + 18 = 0 + 25 [completeing square for x terms and y terms(see
(x+5)^2 + y^2 + 6y + 18 =25 below
(x + 5)^2 + y^2 + 6y + 18 - 9 = 25-9
(x + 5)^2 + y^2 + 6y + 9 = 16
(x + 5)^2 + (y+3)^2 = 16
(x + 5)^2 + (y+3)^2 = 4^2
THIS is in form of circle's equation (x-h) + (y-k) = r^2
where h, k is the centre of the circle and r is the radius. this means that the centre is at (-5,-3) while the radius is root 16, or 4.

Answer 3

u for got it's -6y

Answer 4

not +6y

Answer 5

Robert just copied from a yahoo answer's question, can anyone explain how to do this

Answer 6

@robert2 plz mention link where you are copying
http://answers.yahoo.com/question/index?qid=20111116150842AA4SlZr

Answer 7

Since the normal, easy to look circle formula looks like this:
\( (x-a)^2 + (y-b)^2 = r^2\) where (a,b) is the center,
all you have to do is to rearrange \(x^2 + y^2 + 10x – 6y + 18 = 0\). into this.
Are you familiar with completing the square?