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

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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 th

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

Vedclass · Answer

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

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

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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