Translations of Conics (circles in standard form)

[juliaann1987]

In exercises 7-10, write the equation of the circle in standard form, and then identify its center and radius.

I am having trouble getting number 7 and number 9. I know what the answers are but i don't know how to get to them. If can find a way to figure out the center and the radius i can put the equation in standard form. Any help would be greatly appreciated.

Standard form of circle: (x - h)^2 + (y - k)^2 = r^2 center: (h, k)

7. x^2 + y^2 - 2x + 6y + 9 = 0


9. 4x^2 + 4y^2 + 12x - 24y + 41 = 0

 

[galactus]

I'll step through #1:

\(\displaystyle \L\\x^{2}-2x+y^{2}+6y=-9\)

Complete the squares:

\(\displaystyle \L\\(x^{2}-2x+?)+(y^{2}+6y+ ?)=-9\)

Square half the coefficient of x in each:

\(\displaystyle \L\\(x^{2}-2x+1)+(y^{2}+6y+9)=-9+1+9\)

Factor:

\(\displaystyle \L\\(x-1)^{2}+(y+3)^{2}=1\)

Voila!!

 

[juliaann1987]

You are amazing!!!!! Thank you soooo much.

 

[juliaann1987]

Can the second problem be solved the same way????

 

[galactus]

It sure can. It's a wee bit trickier. I would suggest dividing through by 4 first.

 

[juliaann1987]

Thanks again