SOLUTION: what is the area of the circle 4x^2+4y^2-7x+6y-35=0

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Question 771467: what is the area of the circle 4x^2+4y^2-7x+6y-35=0
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!

We must get the circle in standard form:

      (x - h)² + (y - k)² = r²

in order to get the radius.

 4x² + 4y² - 7x + 6y - 35 = 0

     4x² - 7x  + 4y² + 6y = 35

Divide through by 4

      x² - x  + y² + y = 

      x² - x  + y² + y = 

  (x² - x)  + (y² + y) = 

Complete the square in each parentheses)

For the first parentheses:

Multiply the coefficient of x, which is  by ,
getting , then square that  = 
Add that in the first parentheses and also add it to the right side.

For the second parentheses:

Multiply the coefficient of y, which is  by ,
getting , then square that  = 
Add that in the second parentheses and also add it to the right side.

  (x² - x + )  + (y² + y + ) =  +  + 

Factor each parentheses

  (x - )² + (y - ) =  +  + 

Get a LCD of 64 for the fractions on the right side:
 
    +  +  =  +  +  = 

So the center is (, ) and the radius is  = .  That wasn't necessary because we only need
r²=  to find the area.

The area is 

A = pr² = p() =  which is approximately 31.66 square units.



Edwin


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