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JEE ADVANCE 2010
paper 1
Question
The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
Select the correct option:
A
$2 x-\sqrt{5} y-20=0$
B
$2 x-\sqrt{5} y+4=0$
C
$3 x-4 y+8=0$
D
$4 x-3 y+4=0$
βœ“ Correct! Well done.
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Solution
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