Mathematics · 2021
JEE Main · 16 March 2021, Shift 2 · Q62
Let the lengths of intercepts on x -axis and y -axis made by the circle x^2+y^2+ a x+2 a y+ c =0, ( a <0) be 2 √ 2 and 2 √ 5, respectively. Then the…
Let the lengths of intercepts on $\displaystyle x$-axis and $\displaystyle y$-axis made by the circle $\displaystyle x^2+y^2+\mathbf{a} x+2 \mathbf{a} y+\mathrm{c}=0$, $\displaystyle (\mathrm{a}<0)$ be $\displaystyle 2 \sqrt{2}$ and $\displaystyle 2 \sqrt{5}$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line $\displaystyle x+2 y=0$, is equal to :
Official answer
(4)
$\displaystyle \sqrt{6}$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.