Find the equation of a circle given two tangent lines, and a line passing through center

Question

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:

$x^2+y^2-10x+4y+28=0$

$x^2+y^2-6x+\frac{4}{7}y+\frac{324}{49}=0$

Answer 1

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 $$r=\frac{275}{74},x_0=\frac{1679}{74},y_0=\frac{635}{37}$$

Answer 2

Your equation for the line $z$ is wrong; it should be $6x+7y-16=0$. Your problem becomes$$\left\{\begin{array}{l}\frac{|8 x + 15 y + 7|}{17}=\frac{|3 x - 4 y - 18|}5\\6x+7y=16,\end{array}\right.$$whose solutions are $(5,-2)$ and $\left(3,-\frac27\right)$. Can you take it from here?

Answer 3

Hint:

I would suggest a different approach, noting that the circle will have the center on the lines (there are two) bisecting the angles between $r$ and $s$.

So
- detemine the point (say $Q$) of intersection between $r$ and $s$;
- determine the unit vectors normal to $r$ and $s$ ($\bf n_r=(8,15)/\sqrt{8^2+15^2}$ etc.);
- the normal vectors $\bf b_1,\bf b_2$ to the bisecting lines will simply be $\bf n_r \pm \bf n_s$: don't need to take them unitary;
- having the normal vectors determine the lines normal to them , passing through $Q$;
- $C_1, C_2$ will be their crossing with line $z$;
- the radii will be the (absolute) value of the dot product of $\vec {QC_n}$ with $\bf n_r$ (or alternatively with $\bf n_s$).