This geometry question feels like it should come very easily but alas, it isn't, so I find myself here.
I have the following information: (z) passes through my center, (s) and (r) are my tangent lines.
$$(z) 6x-7y-16=0$$
$$(r) 8x+15y+7=0$$
$$(s) 3x-4y-18=0$$
So given a circle with center ($x_0$,$y_0$) I know the distance between my center and (s) will be equal to the distance between my center and (r).
$d(C,r) = \frac{|8x_0 + 15y_0 + 7|}{\sqrt(8^2 + 15^2)} =\frac{8x_0 + 15y_0 + 7}{17} $
$d(C,s) = \frac{|3x_0 – 4y_0 – 18|}{\sqrt(3^2 + (-4)^2)} =\frac{-3x_0 + 4y_0 + 18}{5}, \frac{3x_0 – 4y_0 – 18}{5} $
$5(8x_0 + 15y_0 + 7) = 17(3x_0 + 4y_0 + 18)$
$40x_0 + 75y_0 + 35 = 51x_0 + 68y_0 + 306$
$y_0 = \frac{11x_0}{7}+\frac{271}{7}$
I substitute this into the equation of the line passing through the center and get $x_0 = 51$. However, even before I continue to operate, I can see this won't give me a correct answer as my book tells me that the two solutions end up being:
$x^2+y^2-10x+4y+28=0$
$x^2+y^2-6x+\frac{4}{7}y+\frac{324}{49}=0$
Best Answer
I think it must be
$$\frac{|8x_0+15y_0+7|}{\sqrt{64+225}}=r$$ $$\frac{|3x_0-4y_0-18|}{\sqrt{9+16}}=r$$
$$6x_0-7y_0-16=0$$ One solution is given by $$r=\frac{275}{74},x_0=\frac{1679}{74},y_0=\frac{635}{37}$$