The equation of the tangent to the circle x^2 | vedclass

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Question

(A) The equation of a line perpendicular to $y=mx+1$ is of the form $x+my+k=0$,or $y=-\frac{1}{m}x+c$. For a circle $x^2+y^2=r^2$,the condition for th

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$x+my \pm \sqrt{1+m^2}=0$

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(A) The equation of a line perpendicular to $y=mx+1$ is of the form $x+my+k=0$,or $y=-\frac{1}{m}x+c$. For a circle $x^2+y^2=r^2$,the condition for the line $y=m_1x+c$ to be a tangent is $c^2=r^2(1+m_1^2)$. Here,the slope of the tangent is $m_1 = -\frac{1}{m}$ and $r=1$. Substituting these into the condition: $c^2 = 1 \cdot (1 + (-\frac{1}{m})^2) = 1 + \frac{1}{m^2} = \frac{m^2+1}{m^2}$. Thus,$c = \pm \frac{\sqrt{m^2+1}}{m}$. The equation of the tangent is $y = -\frac{1}{m}x \pm \frac{\sqrt{m^2+1}}{m}$. Multiplying by $m$,we get $my = -x \pm \sqrt{m^2+1}$,which simplifies to $x+my \pm \sqrt{1+m^2}=0$.

$List-I$	$List-II$
$(P) \alpha \text{ equals}$	$(1) (-2,4)$
$(Q) r \text{ equals}$	$(2) \sqrt{5}$
$(R) A_1 \text{ equals}$	$(3) (-2,6)$
$(S) B_1 \text{ equals}$	$(4) 5$
	$(5) (2,4)$

The equation of the tangent to the circle $x^2+y^2=1$,which is perpendicular to the line $y=mx+1$,is:

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