Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

12,143 questions

16,810 answers

1,423 comments

10,725 users

What is the radius of a circle with the equation x^2+y^2+6x-2y+3=0?

0 votes

Round the answer to the nearest thousandth?

asked Jan 11, 2013 in PRECALCULUS by linda Scholar

1 Answer

+3 votes

x^2+y^2+6x-2y+3 = 0

⇒x^2+y^2+6x-2y+3 = 0

Add 9 to each side

⇒x^2+y^2+6x-2y+3 + 9 = 9

⇒x^2+6x+9+y^2-2y+3 = 9

Add 1 to each side

⇒x^2+6x+9+y^2-2y+3 +1= 9+1

⇒(x^2+6x+9)+(y^2-2y+1)+3 = 9+1

Now it cnn be written as [a^2+2ab+b^2 = (a+b)^2 , a^2-2ab+b^2 = (a-b)^2 ]

(x+3)^2+(y-1)^2+3 = 10

Subtract 3 from each side.

(x+3)^2+(y-1)^2+3 - 3 = 10 - 3

Simplify

(x+3)^2+(y-1)^2 = 7

This equations compered to (x-h)^2+(y-k)^2 = r^2 [ center (-h , -k)    radius = r^2

Here h = -3 , k = 1 and r^2 = 7

There fore

Center ( -3 , 1)

Radius r^2 = 7

r = √7

= 2.646

There fore  Center ( -3 , 1)  , r = 2.646

answered Jan 11, 2013 by richardson Scholar

Related questions

...