If the chord L equiv y-mx-1=0 of the circle S | vedclass

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(B) The line $L \equiv -mx+y-1=0$ is a chord of the circle $S \equiv x^2+y^2-1=0$ and it touches the circle $S_1 \equiv x^2+y^2-4x+1=0$.The center of

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$(2 \pm \sqrt{6}, 1)$

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(B) The line $L \equiv -mx+y-1=0$ is a chord of the circle $S \equiv x^2+y^2-1=0$ and it touches the circle $S_1 \equiv x^2+y^2-4x+1=0$.The center of $S_1$ is $(2, 0)$ and its radius is $r = \sqrt{2^2+0^2-1} = \sqrt{3}$.Since the line touches the circle $S_1$,the perpendicular distance from the center $(2, 0)$ to the line $L$ must equal the radius $\sqrt{3}$.$\frac{|-m(2) + 0 - 1|}{\sqrt{(-m)^2 + 1^2}} = \sqrt{3}$$\frac{|-2m-1|}{\sqrt{m^2+1}} = \sqrt{3}$Squaring both sides: $\frac{(2m+1)^2}{m^2+1} = 3$$4m^2 + 4m + 1 = 3m^2 + 3$$m^2 + 4m - 2 = 0$Solving for $m$: $m = \frac{-4 \pm \sqrt{16 - 4(1)(-2)}}{2} = \frac{-4 \pm \sqrt{24}}{2} = -2 \pm \sqrt{6}$.Substituting $m$ into $L$: $y - (-2 \pm \sqrt{6})x - 1 = 0$,which simplifies to $y + (2 \mp \sqrt{6})x - 1 = 0$.The chord of contact of a point $(h, k)$ with respect to $S \equiv x^2+y^2-1=0$ is $hx + ky - 1 = 0$.Comparing $hx + ky - 1 = 0$ with $(2 \mp \sqrt{6})x + y - 1 = 0$,we get $h = 2 \mp \sqrt{6}$ and $k = 1$.Thus,the point is $(2 \pm \sqrt{6}, 1)$.

If the chord $L \equiv y-mx-1=0$ of the circle $S \equiv x^2+y^2-1=0$ touches the circle $S_1 \equiv x^2+y^2-4x+1=0$,then the possible points for which $L=0$ is a chord of contact of $S=0$ are

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