If a line, y=m x+c is a tangent to circle, (x-3)^{2}+y^{2}=1 and it is perpendicular to a line mathr

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10-1.Circle and System of Circles

hard

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

A

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

B

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

C

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

D

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

(JEE MAIN-2020)

Solution

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 $(x-3)^{2}+y^{2}=1$

$\left|\frac{c+3}{\sqrt{2}}\right|=1 \Rightarrow c^{2}+6 c+7=0$

Standard 11

Mathematics

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