SOLUTION: Dear Sir/Madam, I am confronted with the following problem: "Find the center and radius of the circle with the following equation: x^2 + y^2 - 10x + 8y -40 = 0" I have

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Dear Sir/Madam, I am confronted with the following problem: "Find the center and radius of the circle with the following equation: x^2 + y^2 - 10x + 8y -40 = 0" I have      Log On


   



Question 6644: Dear Sir/Madam,
I am confronted with the following problem:
"Find the center and radius of the circle with the following equation:
x^2 + y^2 - 10x + 8y -40 = 0"
I have absolutely no idea how to approach this question. I want to use the x^2 + y^2 = r^2 formula, but I don't how to integrate that into this problem. Can you please help me?
Thanks in advance.
Regards,
-Mike

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
This is a completing the square problem:
x%5E2+%2B+y%5E2+-+10x+%2B+8y+-40+=+0

Get the x terms together, and the y terms together, and leave a space after each to complete the square. Also add + 40 to each side of the equation:
x%5E2+-+10x+%2B+____+%2B+y%5E2+%2B+8y+%2B+____+=+40+%2B+____+%2B+____+

For the first blank, take half of the 10, which is 5 and square to get 25.
For the second blank, take half of the 8, which is 4 and square to get 16. Add these to both sides of the equation:
x%5E2+-+10x+%2B+25+%2B+y%5E2+%2B+8y+%2B+16+=+40+%2B+25+%2B+16+
+%28x-5%29%5E2+%2B+%28y%2B4%29%5E2+=+81

Therefore this is a circle whose center is at (5, -4) and r%5E2+=+81 so r=9.

R^2 at SCC