# center and radius of a circle?

• Jan 29th 2009, 09:24 PM
lsnyder
center and radius of a circle?
how do you find the center and radius of a circle?

Problem:
Find the center of the circle given in general form (x^2)+(y^2)-10x-2y+50=49
• Jan 29th 2009, 09:30 PM
Chris L T521
Quote:

Originally Posted by lsnyder
how do you find the center and radius of a circle?

Problem:
Find the center of the circle given in general form (x^2)+(y^2)-10x-2y+50=49

Hint: $\displaystyle x^2+y^2-10x-2y+50=49\implies x^2+y^2-10x-2y+25+1+24=49$

Group the proper terms together, such that you generate perfect squares.

Can you take it from here?
• Jan 29th 2009, 09:35 PM
lsnyder
Quote:

Originally Posted by Chris L T521
Hint: $\displaystyle x^2+y^2-10x-2y+50=49\implies x^2+y^2-10x-2y+25+4+21=49$

Group the proper terms together, such that you generate perfect squares.

Can you take it from here?

i don't fully understand, sorry...........
• Jan 29th 2009, 09:39 PM
Chris L T521
Quote:

Originally Posted by lsnyder
i don't fully understand, sorry...........

I'll show the next step.

You should group the terms as follows : $\displaystyle \left(x^2-10x+25\right)+\left(y^2-2y+1\right)=49-24$

The terms in parenthesis on the LHS of the equation factor as perfect squares. On the right side, you'll have the radius value squared.

Can you continue from here?

EDIT: Please refer to my first post. I fixed an error I found in it.
• Jan 29th 2009, 09:46 PM
lsnyder
Quote:

Originally Posted by Chris L T521
I'll show the next step.

You should group the terms as follows : $\displaystyle \left(x^2-10x+25\right)+\left(y^2-2y+1\right)=49-24$

The terms in parenthesis on the LHS of the equation factor as perfect squares. On the right side, you'll have the radius value squared.

Can you continue from here?

EDIT: Please refer to my first post. I fixed an error I found in it.

i understand now, thank you very much!