# Find circle known tangent line

circle $$x^2 + y^2 + Ax + By + C = 0$$ has tangent line of $$y = -2x -6$$ and $$y=0.5x +4$$ also point (-1,6) on the circle. What the possible $$y$$ coordinate of its center? I draw the plane as

Lets say circle and $$y = -2x -6$$ intersect at $$(x_1, y_1)$$ It means $$\frac {y_1 - b}{x_1 - a} = \frac 12$$

Circle and $$y=0.5x +4$$ intersect at $$(x_2, y_2)$$ It means $$\frac {y_2 - b}{x_2 - a} = -2$$

Or maybe input $$(x_1, y_1)$$ $$(x_2, y_2)$$ and (-1,6) into $$x^2 + y^2 + Ax + By + C = 0$$.

From center (a,b) to (-1,6) is radius. From $$(x_1, y_1)$$ to (a,b) also radius. From (a,b) to $$(x_2, y_2)$$ also radius.

Maybe finding the tangent line known the circle is easier than finding the circle known tangent line... how do i find the center of the circle?

Let $$E(a,b).$$
Thus, $$\sqrt{(a+1)^2+(b-6)^2}=\frac{|2a+b+6|}{\sqrt5}=\frac{|a-2b+8|}{\sqrt5}.$$ Can you end it now?
I got $$E(1,17)$$ or $$E(-3,5).$$