



 
Hi Ryan, I didn't get the same answer as you did so I want to show you the process with a different problem.
To see the centre and the radius I want to write the circle in the form (x  a)^{2} + (y  b)^{2} = r^{2} for then I know that the circle has centre (a, b) and radius r. For the moment look at (x  a)^{2} = x^{2}  2ax + a^{2}. In my problem the terms containing x are x^{2} and  6x and comparing the xterms, ^{}  2ax and 6x^{}, a must be 3. That is a is 6/2. Similarly (y  b)^{2} = y^{2}  2by + b^{2} and comparing 2by to 12y, b must be 6. Here b is 12/2. So a is minus half the coefficient of x and b is minus half the coefficient of y. So here is my solution to
Hence the centre is (3, 6) and the radius is √39. Now try your problem again. Ryan wrote back hello I took the advice on the problem to find the center of the circle and the radious the prob is this x^2+y^2+8x2y+15=0 is this correct Hi again Ryan, I put the equation of the circle in standard form and got
Thus the centre is (4, 1) but the radius is √2. Harley  


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