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JEE Mains · Maths · STD 11 - 10.1 circle and system of 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 \(\mathrm{L}_{1},\) 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

  1. A \(c^{2}-6 c+7=0\)
  2. B \(c^{2}+6 c+7=0\)
  3. C \(c^{2}+7 c+6=0\)
  4. D \(c^{2}-7 c+6=0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(c^{2}+6 c+7=0\)

Step-by-step Solution

Detailed explanation

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 \((x-3)^{2}+y^{2}=1\)…
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