Circle $x^{2} + y^{2} - 4 x - 8 y - 5 = 0$ will intersect the line $3 x - 4 y = m$ in two distinct points, if

- 10 < m < 5

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9 < m < 20

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- 35 < m < 15

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Solution

The correct option is C $- 35 < m < 15$

We have,
$x^{2} + y^{2} - 4 x - 8 y - 5 = 0 {(x - 2)}^{2} + {(y - 4)}^{2} = {(5)}^{2} \frac{| 3 (2) - 4 (4) - m |}{5} < 5 = | 6 - 16 - m | < 25 = | m + 10 | < 25 - 25 < m + 10 < 25 - 35 < m < 15$

Hence, this is the answer.

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The circle $x^{2} + y^{2} = 4 x + 8 y + 5$ interesects the line 3x - 4y = m at two distinct points if
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The circle $x^{2} + y^{2} = 4 x + 8 y + 5$ interesects the line 3x - 4y = m at two distinct points if
(2010)

3 x - 4 y = m

x^{2} + y^{2} - 4 x - 8 y - 5 = 0

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(x^{2} + y^{2} - 4 x - 8 y - 5 = 0)

λ

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