JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 10)
Let the circle S : 36x2 + 36y2 $$-$$ 108x + 120y + C = 0 be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, x $$-$$ 2y = 4 and 2x $$-$$ y = 5 lies inside the circle S, then :
$${{25} \over 9} < C < {{13} \over 3}$$
100 < C < 165
81 < C < 156
100 < C < 156
Explanation
S : 36x2 + 36y2 $$-$$ 108x + 120y + C = 0
$$\Rightarrow$$ x2 + y2 $$-$$ 3x + $${{10} \over 3}$$y + $${C \over {36}}$$ = 0
Centre $$ \equiv ( - g, - f) \equiv \left( {{3 \over 2},{{ - 10} \over 6}} \right)$$
radius = $$r = \sqrt {{9 \over 4} + {{100} \over {36}} - {C \over {36}}} $$
_22th_July_Evening_Shift_en_10_1.png)
Now,
$$ \Rightarrow r < {3 \over 2}$$
$$ \Rightarrow {9 \over 4} + {{100} \over {36}} - {C \over {36}} < {9 \over 4}$$
$$\Rightarrow$$ C > 100 ...... (1)
Now, point of intersection of x $$-$$ 2y = 4 and 2x $$-$$ y = 5 is (2, $$-$$1), which lies inside the circle S.
$$\therefore$$ S(2, $$-$$1) < 0
$$\Rightarrow$$ (2)2 + ($$-$$1)2 $$-$$ 3(2) + $${{10} \over 3}$$($$-$$1) + $${C \over {36}}$$ < 0
$$\Rightarrow$$ 4 + 1 $$-$$ 6 $$-$$ $${{10} \over 3}$$ + $${C \over {36}}$$ < 0
C < 156 ..... (2)
From (1) & (2)
100 < C < 156 Ans.
$$\Rightarrow$$ x2 + y2 $$-$$ 3x + $${{10} \over 3}$$y + $${C \over {36}}$$ = 0
Centre $$ \equiv ( - g, - f) \equiv \left( {{3 \over 2},{{ - 10} \over 6}} \right)$$
radius = $$r = \sqrt {{9 \over 4} + {{100} \over {36}} - {C \over {36}}} $$
Now,
$$ \Rightarrow r < {3 \over 2}$$
$$ \Rightarrow {9 \over 4} + {{100} \over {36}} - {C \over {36}} < {9 \over 4}$$
$$\Rightarrow$$ C > 100 ...... (1)
Now, point of intersection of x $$-$$ 2y = 4 and 2x $$-$$ y = 5 is (2, $$-$$1), which lies inside the circle S.
$$\therefore$$ S(2, $$-$$1) < 0
$$\Rightarrow$$ (2)2 + ($$-$$1)2 $$-$$ 3(2) + $${{10} \over 3}$$($$-$$1) + $${C \over {36}}$$ < 0
$$\Rightarrow$$ 4 + 1 $$-$$ 6 $$-$$ $${{10} \over 3}$$ + $${C \over {36}}$$ < 0
C < 156 ..... (2)
From (1) & (2)
100 < C < 156 Ans.
Comments (0)
