<s>[INST] Co-ordinate Geometry : Circle Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, \frac{1}{2})$. Find the slope of the line.
Comments: First I write the family of lines passing through the given point i.e., $y - \frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 \sqrt{1 + m^2}$ but the solution of this equation is lengthy.
I am seeking a shorter solution.
 [/INST] *

*Line $$2mx-2y+4m+1=0 \tag{1}$$

*First circle $$x^2+y^2=1$$
centre $O=(0,0)$, radius $r_1=1$
distance of $O$ from $(1)$: $$d_1=\left| \frac{4m+1}{2\sqrt{m^2+1}} \right|$$

*Second circle $$x^2+y^2-8x+11=0$$
centre $P=(4,0)$, radius $r_2=\sqrt{4^2-11}=\sqrt{5}$
distance of $P$ from $(1)$: $$d_2=\left| \frac{12m+1}{2\sqrt{m^2+1}} \right|$$

*Equating semi-chord length: $r^2-d^2$
\begin{align}
   r_1^2-d_1^2 &= r_2^2-d_2^2 \\
   1-\frac{(4m+1)^2}{4(m^2+1)} &=
   5-\frac{(12m+1)^2}{4(m^2+1)} \\
   m &= \frac{-1\pm \sqrt{29}}{14}
 \end{align}
Only the negative root gives intersections:

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