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Mathematics · 2020
JEE Main · 7 January 2020, Shift 2 · Q64
Let the tangents drawn from the origin to the circle, x^2+y^2-8 x-4 y+16=0 touch it at the points A and B. The ( AB )^2 is equal to:
Let the tangents drawn from the origin to the circle, $\displaystyle x^2+y^2-8 x-4 y+16=0$ touch it at the points A and B . The $\displaystyle (\mathrm{AB})^2$ is equal to :
Official answer
(3)
$\displaystyle \frac{64}{5}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.