SOLUTION: Please help me, I am lost. complete the square and write the equation in standard form. Then give the center and radius of the circle: x^2+y^2-2x-4y-4=0 Thank you so much

Question 134905: Please help me, I am lost.
complete the square and write the equation in standard form. Then give the center and radius of the circle:
x^2+y^2-2x-4y-4=0
Thank you so much
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Step 1: Set the constant term on the right side of the equation:

Step 2: Rearrange the terms, put the x terms together and the y terms together:

Step 3: Divide the coefficient on the x term by 2, and then square the result:

Step 4: Add this value to both sides of the equation:

Step 5: Repeat steps 3 and 4 using the coefficient on the y term:

is a perfect square, so

is a perfect square, so

Step 6: Using this information, re-write the equation:

This is now in the standard form of an equation of a circle with center at (h,k) and radius r, namely

Step 7: You can now tell by inspection that the x-coordinate of the center is 1, the y-coordinate of the center is 2, and the radius is .

C(1,2), r = 3
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