Question

Standard Equation of a Circle: ax2 + by2 + cx + dy + e = 0
How do I find the center and radius of this circle when a and b do not equal one. (I understand that a and b must be equal to form a true circle.) This needs to be algebraic (not numbers), so that we can plug in any data set into it.

Answer 1

\[ax^2+by^2+cx+dy+e=0\]

Answer 2

Think about it like this:
(x-a)² + (y-b)² = r², where (a,b) is the center, and r the radius - so, you'd put in 3 = r, and (0,b) where b could be any number:
x² + (y-b)² = 9 for the circle in your question, then expand it into the general form of your question:
= x² + y² - 2by + b² = 9
= 1x² + 1y² +0x + (-2b)y + (b²-9) = 0, and then read off A, B, C, D, and E from that:
A=1, B=1, C=0, D = -2b, and E = b²-9.

Answer 3

divide by \(a\) then write in standard form

Answer 4

No one is asking you to assume that a = b = 1.

Divide the entire equation by the coefficient 'a,' and then write in the resulting equation in standard form. Will you need to use "completing the square" or not?

Answer 5

That actually makes more sense. Thanks. I will be completing the square.