SOLUTION: The circle with equation x2 + y2 - 2x + 4y = 0 has centre C and radius r; where 1. C (1;-2) ; r = square root of 5 2.C (-1; 2) ; r = square root of 5 3. C (1;-4) ; r = square ro

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Question 1082829: The circle with equation x2 + y2 - 2x + 4y = 0 has centre C and radius r; where
1. C (1;-2) ; r = square root of 5
2.C (-1; 2) ; r = square root of 5
3. C (1;-4) ; r = square root of 17
4.C (-1; 4) ; r = 17
5. C (1;-2) ; r = 5
please assist with this one
Answer by Edwin McCravy(20059)   (Show Source): You can put this solution on YOUR website!
Instead of doing yours for you, I will do another
one exactly like yours, which you can use as a model 
to do yours by:



Rearrange terms so that x terms are together and 
y terms are together.

 

We insert blanks where we must insert numbers
which complete the squares:



We complete the square to find what goes in the first 
blanks on each side of the equation.

1. Multiply the coefficient of x by 1/2.
    (-6)(1/2) = -3
2. Square -3, get (-3)² = +9
3. Add to both sides in the first blanks on each
   side.



We also complete the square to find what goes in the 
remaining blanks on each side of the equation.

1. Multiply the coefficient of y by 1/2.
    (+8)(1/2) = +4
2. Square +4, get (+4)² = +16
3. Add to both sides in the remaining blanks on each
   side.



Next we factorise the first three terms on the left
as (x-3)² and the last three terms on the left as (y+4)²
and combine terms on the right as 26



We compare that to the standard equation for a circle:


                      __
and h=3, k=-4, and r=√26
                                   __
So centre is (3,-4), radius = r = √26 

Now you can do yours exactly the same way.

Edwin
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