JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 14)
Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept $$6\sqrt 5 $$ on the x-axis. Then the radius of the circle C is equal to :
$$\sqrt {53} $$
9
8
$$\sqrt {82} $$
Explanation
_27th_July_Evening_Shift_en_14_1.png)
$$r = \sqrt {{6^2} + {{\left( {3 + \sqrt 5 } \right)}^2}} $$
$$ = \sqrt {36 + 45} = 9$$
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