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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

Slope of tangent to $\mathrm{x}^{2}+\mathrm{y}^{2}=1$ at $\mathrm{P}\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

$2 \mathrm{x}+2 \mathrm{yy}^{\prime}=\left.0 \Rightarrow \mathrm{m}_{\mathrm{T}}\right|_{\mathrm{P}}=-1$

$y=m x+c$ is tangent to $(x-3)^{2}+y^{2}=1$

$y=x+c$ is tangent to $(x-3)^{2}+y^{2}=1$

$\left|\frac{c+3}{\sqrt{2}}\right|=1 \Rightarrow c^{2}+6 c+7=0$

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 $\mathrm{L}_{1},$ where $\mathrm{L}_{1}$ is the tangent to the circle, $\mathrm{x}^{2}+\mathrm{y}^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ then

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