JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 6)
A square is inscribed in the circle x2 + y2
– 6x + 8y – 103 = 0 with its sides parallel to the coordinate axes.
Then the distance of the vertex of this square which is nearest to the origin is :
$$\sqrt {137} $$
6
$$\sqrt {41} $$
13
Explanation
_11th_January_Morning_Slot_en_6_1.png)
R $$ = \sqrt {9 + 16 + 103} = 8\sqrt 2 $$
OA $$ = 13$$
OB $$ = \sqrt {265} $$
OC $$ = \sqrt {137} $$
OD $$ = \sqrt {41} $$
Comments (0)
