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 $E(-3,5).$