how to solve 7x^2+7y^2-28x+63y-4=0 in general equation of circle?
not sure what you mean by "solve"
If you want to find the center and radius, complete the squares:
7x^2+7y^2-28x+63y-4 = 0
7(x^2-4x+4) + 7(y^2+9y+(9/2)^2) = 4 + 7*4 + 7(9/2)^2
7(x-2)^2 + 7(y + 9/2)^2 = 695/4
(x-2)^2 + (y + 9/2)^2 = 695/28
now you can read off the center and radius
To solve the equation 7x^2 + 7y^2 - 28x + 63y - 4 = 0 in the general equation of a circle, we need to complete the square.
Step 1: Rearrange the equation
Start by rearranging the original equation to group the x-terms, y-terms, and constant term together:
7x^2 - 28x + 7y^2 + 63y = 4
Step 2: Complete the square for x-terms
To complete the square for the x-terms, take half of the coefficient of x (which is -28/2 = -14) and square it (-14^2 = 196):
7(x^2 - 4x) + 7y^2 + 63y = 4 + 196
Step 3: Complete the square for y-terms
To complete the square for the y-terms, take half of the coefficient of y (which is 63/2 = 31.5) and square it (31.5^2 = 992.25):
7(x^2 - 4x) + 7(y^2 + 63y) = 4 + 196 + 992.25
Step 4: Simplify the equation
Simplify the equation by combining the constants on the right side of the equation:
7(x^2 - 4x) + 7(y^2 + 63y) = 1192.25
Step 5: Factor out the common coefficient
Factor out the common coefficient of 7 from the x-terms and y-terms:
7(x^2 - 4x + ____ ) + 7(y^2 + 63y + ____ ) = 1192.25
Step 6: Complete the square for x-terms and y-terms individually
To complete the square for the x-terms, take half of the coefficient of x (which is -4/2 = -2) and square it (-2^2 = 4). Similarly, for the y-terms, take half of the coefficient of y (which is 63/2 = 31.5) and square it (31.5^2 = 992.25):
7(x^2 - 4x + 4 ) + 7(y^2 + 63y + 992.25 ) = 1192.25 + 28 + 6945.75
Step 7: Simplify the equation further
Simplify the equation by combining the constants on the right side:
7(x - 2)^2 + 7(y + 31.5)^2 = 8136
The equation is now in the general equation of a circle form: (x - h)^2 + (y - k)^2 = r^2, where the center of the circle is at the point (h, k) and the radius is r.
In this case, the center of the circle is at (2, -31.5) and the radius is √(8136/7).