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) About Me  (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 - 7%2F4x  + y² + 6%2F4y = 35%2F4

      x² - 7%2F4x  + y² + 3%2F2y = 35%2F4

  (x² - 7%2F4x)  + (y² + 3%2F2y) = 35%2F4

Complete the square in each parentheses)

For the first parentheses:

Multiply the coefficient of x, which is -7%2F4 by 1%2F2,
getting -7%2F8, then square that %28-7%2F8%29%5E2 = 49%2F64
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 3%2F2 by 1%2F2,
getting 3%2F4, then square that %283%2F4%29%5E2 = 9%2F16
Add that in the second parentheses and also add it to the right side.

  (x² - 7%2F4x + 49%2F64)  + (y² + 3%2F2y + 9%2F16) = 35%2F4 + 49%2F64 + 9%2F16

Factor each parentheses

  (x - 7%2F8)² + (y - 3%2F4) = 35%2F4 + 49%2F64 + 9%2F16

Get a LCD of 64 for the fractions on the right side:
 
   35%2F4 + 49%2F64 + 9%2F16 = 560%2F64 + 49%2F64 + 36%2F64 = 645%2F64

So the center is (7%2F8, 3%2F4) and the radius is sqrt%28645%2F64%29 = sqrt%28645%29%2F8.  That wasn't necessary because we only need
r²= 645%2F64 to find the area.

The area is 

A = pr² = p(645%2F64) = %28645pi%29%2F64 which is approximately 31.66 square units.



Edwin