If a line,y=mx+c is a tangent to the circle,( | vedclass

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(B) The equation of the circle is $x^{2}+y^{2}=1$. The slope of the tangent at $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ is found by diff

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$c^{2}+6c+7=0$

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(B) The equation of the circle is $x^{2}+y^{2}=1$. The slope of the tangent at $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ is found by differentiating: $2x+2yy'=0 \Rightarrow y' = -\frac{x}{y}$.At $P$,the slope $m_{L1} = -\frac{1/\sqrt{2}}{1/\sqrt{2}} = -1$.Since the line $y=mx+c$ is perpendicular to $L_{1}$,its slope $m = -\frac{1}{m_{L1}} = -\frac{1}{-1} = 1$.Thus,the line is $y=x+c$,or $x-y+c=0$.This line is a tangent to the circle $(x-3)^{2}+y^{2}=1$,which has center $(3, 0)$ and radius $r=1$.The perpendicular distance from the center $(3, 0)$ to the line $x-y+c=0$ must equal the radius:$\frac{|3-0+c|}{\sqrt{1^{2}+(-1)^{2}}} = 1 \Rightarrow |3+c| = \sqrt{2}$.Squaring both sides: $(3+c)^{2} = 2$ $\Rightarrow 9+6c+c^{2} = 2$ $\Rightarrow c^{2}+6c+7=0$.

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

If a line,$y=mx+c$ is a tangent to the circle,$(x-3)^{2}+y^{2}=1$ and it is perpendicular to a line $L_{1},$ where $L_{1}$ is the tangent to the circle,$x^{2}+y^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ then

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