SOLUTION: if x+y^2+kxy+8x-6y+9=0 represents a circle state the value of k and substituting this value of k in the equation;find the center and radius of the circle

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: if x+y^2+kxy+8x-6y+9=0 represents a circle state the value of k and substituting this value of k in the equation;find the center and radius of the circle      Log On


   

Question 697821: if x+y^2+kxy+8x-6y+9=0 represents a circle state the value of k and substituting this value of k in the equation;find the center and radius of the circle
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
It should have a term with x%5E2.
If x%5E2%2By%5E2%2Bkxy%2B8x-6y%2B9=0 represents a circle,
k=0.
(Terms in xy appear when a conic section other than a circle has axes that are not parallel to the x- and y-axes, but circles do not care about rotating the axes).

x%5E2%2By%5E2%2B8x-6y%2B9=0 --> %28x%5E2%2B8x%29%2B%28y%5E2-6y%29=-9 --> %28x%5E2%2B8x%2B16%29%2B%28y%5E2-6y%2B9%29=-9%2B16%2B9 --> %28x%2B4%29%5E2%2B%28x-3%29%5E2=16 --> %28x%2B4%29%5E2%2B%28x-3%29%5E2=4%5E2
The last equation shows that the circle is centered at the point (-4,3),
because x%5BC%5D=-4 and y%5BC%5D=3 are the coordinates of the center, subtracted from x and y in the equation.
It also shows that the radius is 4.