x^2+y^2-6x-2y+8=0
Completing the squares results in
(x-3)^2 + (y-1)^2 = 2 so the circle center is at 3,1 and the radius is sqrt(2) as you found
tangent line passing through (0,0) ? There are TWO of them....
Since the line passes through 0,0 the qaution of the line will be of the form y= kx (b will be zero)
Substitute this into the circle equation:
(x-3)^2 + (kx-1)^2 =2 and simplify to :
x^2(1+k^2) + x(6+2k) + 8 =0 (Form: ax^2 + bx +c =0)
we want b^2 = 4ac substituting and using the quadratic formula results in k = 1 and -1/7
so your two tangent lines passing through 0,0 will be
y = x and y=-1/7x
