If a line, y=m x+c is a tangent to the circle,


If a line, $y=m x+c$ 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:

  1. (1) $c^{2}-7 c+6=0$

  2. (2) $c^{2}+7 c+6=0$

  3. (3) $c^{2}+6 c+7=0$

  4. (4) $c^{2}-6 c+7=0$

Correct Option: , 3


Slope of tangent of $x^{2}+y^{2}=1$ at $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

$\frac{1}{\sqrt{2}} x+\frac{1}{\sqrt{2}} y-1=0$

$x+y \sqrt{2}=0$, which is perpendicular to $x-y+c=0$

At $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ which is tangent of $(x-3)^{2}+y^{2}=1$

So, $m=1 \Rightarrow y=x+c$

Now, distance of $(3,0)$ from $y=x+c$ is


$\Rightarrow \quad c=-3 \pm \sqrt{2}$

$\Rightarrow \quad(c+3)^{2}=2$

$\Rightarrow \quad c^{2}+6 c+9=2$

$\therefore \quad c^{2}+6 c+7=0$

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