Mathematics · 2020
JEE Main · 8 January 2020, Shift 2 · Q64
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,…
If a line, $\displaystyle y=\mathrm{m} x+\mathrm{c}$ is a tangent to the circle, $\displaystyle (x-3)^2+y^2=1$ and it is perpendicular to a line $\displaystyle \mathrm{L}_1$, where $\displaystyle \mathrm{L}_1$ is the tangent to the circle, $\displaystyle x^2+y^2=1$ at the point $\displaystyle \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$; then :
Official answer
(3)
$\displaystyle c^2+6 c+7=0$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.