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:

Option 1 -

81 < C < 156

Option 2 -

100 < C < 165

Option 3 -

100 < C < 156

Option 4 -

$\frac{25}{9} < C < \frac{13}{3}$

6 Views | Posted a month ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

$36 x^{2} + 36 y^{2} - 108 x + 120 y + c = 0$

$\Rightarrow x^{2} + y^{2} - 3 x + \frac{10}{3} y + \frac{c}{36} = 0$

The circle not interesting either axes

$∴ g^{2} < c & f^{2} < c$

$\Rightarrow \frac{9}{4} < \frac{c}{36} \Rightarrow c > 81 - (i) & \frac{25}{9} < \frac{c}{36} \Rightarrow c > 100 . . . . . . (i i)$

(i) $\cap$ (ii) -> c > 100 .(iii)

Point of intersection between x – 2y – 4 = 0 & 2x – y – 5 = 0 is (2 - 1) lies inside the circle.

$∴ 4 + 1 - 6 - \frac{10}{3} + \frac{c}{36} < 0 \Rightarrow c < 156 . . . . . . . (i v)$

-> $(i i i) \cap (i v)$ 100 < c < 156

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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:

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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:

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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: