10-1.Circle and System of Circles
hard

જો રેખા $y=m x+c$ એ વર્તુળ $(x-3)^{2}+y^{2}=1$ નો સ્પર્શક છે અને તે રેખા $\mathrm{L}_{1},$ ને લંબ છે કે જ્યાં રેખા $\mathrm{L}_{1}$ એ વર્તુળ $\mathrm{x}^{2}+\mathrm{y}^{2}=1$ નો બિંદુ $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ આગળનો સ્પર્શક હોય તો  . .. .

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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