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Question

Write the equation of the circle, x^2 + y^2 - 4x + 2y - 7 = 0, in standard form.
I know that the formula is r = √(x-h)^2 + (y - k)^2 but i get confused with the values of center and all, please help. Thank you!

Answer 1

Write the equation of the circle, x^2 + y^2 - 4x + 2y - 7 = 0, in standard form.
I know that the formula is r = √(x-h)^2 + (y - k)^2 but i get confused with the values of center and all, please help. Thank you!

 

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Don't be ... First  combine like terms, and try completing squares...

 

   (x^2-4x+4)-4+(y^2+2y+1)-1-7=0    , or

 

   (x-2)^2 + (y+1)^2 = 12   , You're almost there: finally

 

   sqrt((x-2)^2 + (y+1)^2) = sqrt(12)

 

Just double check the arithmetics ...

 

Good luck:)

Answer 2

First let's rearrange the terms.

 

x^2 - 4x + y^2 + 2y - 7 = 0

 

Move the -7 to the right side of the equation.

 

x^2 - 4x + y^2 + 2y = 7

 

To make the above equation look like (x - h)^2 + (y - k)^2 = r^2 you'll need to complete the square.

 

Let's start with the x terms. You have x^2 - 4x. To complete the square, divide the -4 by 2 to get -2, then square it to get +4. Add 4 to each side.

 

x^2 - 4x + 4 + y^2 + 2y = 7 + 4

 

This becomes:

 

(x - 2)^2 + y^2 + 2y = 11

 

Now for the y terms. Take the +2, divide by 2 to get 1, then square it to get +1. Add this to each side.

 

(x - 2)^2 + y^2 + 2y + 1 = 11 + 1

 

This becomes:

 

(x - 2)^2 + (y + 1)^2 = 12

 

Now you're in standard form.

 

Hope this helps.