Circle and Radius, having to complete the square?

**Chaim** · September 11th 2012, 03:48 PM

Hi, I have this problem where I am supposed to find the circle and radius.
Here is what I did and the steps (click on the picture to enlarge it)

Circle and Radius, having to complete the square?-circle-radius-problem.png

And I'm stuck at this part.
I think I am supposed to be using 'completing the square' here right?
But can someone explain to me how to do that?

Because after that, I'm able to find the center, which is (h,k)
Then the radius would be the square root of the number that comes at the other end.

Thanks

**HallsofIvy** · September 11th 2012, 05:01 PM

Yes, complete the square. $4x^2+ 8x= 4(x^2+ 2x)$ . Now what do you have to add to make [itex]x^2+ 2x+ ?[/itex] a perfect square. You should already know that $(x+ a)= x^2+ 2ax+ a^2$ . So if 2a= 2, what is a? And then what is $a^2$ ?

Similarly, $4y^2+ 24y= 4(y^2+ 6y)$ . Now compare $y^2+ 6y$ to $y^2+ 2ay+ a^2$ . If 2a= 6, what is a? What is $a^2$ ?

**skeeter** · September 11th 2012, 05:02 PM

$4x^2+4y^2+8x=-24y+39$

divide every term by 4 ...

$x^2+y^2+2x = -6y + \frac{39}{4}$

re-arrange ...

$x^2+2x+y^2+6y= \frac{39}{4}$

complete the square for x and y ...

$x^2+2x+1+y^2+6y+9= \frac{39}{4}+10$

$(x+1)^2 + (y+3)^2 = \frac{79}{4}$

**Chaim** · September 11th 2012, 05:54 PM

Ah I see ok thanks!
So basically it's like the tic tac toe method (if you guys heard of it), where you basically find numbers that would fit in.

Ok thanks!

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