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Let a circle C of radius 5 lie below the x-axis. The line $L_{1} = 4 x + 3 y + 2 = 0$ passes through the centre P of the circle C and intersects the line $L_{2} : 3 x - 4 y - 11 = 0$ at Q. The line L₂ touches C at the point Q. Then the distance of P from the line 5x-12y+51=0 is

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answer is 11.

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Detailed Solution

Given 4x+3y+2=0 and 3x-4y-11=0 Intersection point Q of these two lines is (1,-2).

$\frac{x}{- 25} = \frac{y}{50} = \frac{1}{- 25}$

$\frac{x - 1}{\cos θ} = \frac{y + 2}{\sin θ} = \pm 5$

$y = - 2 + 5 (- \frac{4}{5}) = - 6, x = 1 + 5 (\frac{3}{5}) = 4$

$P (4, - 6)$

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$Now distance of P from 5 x - 12 y + 51 = 0 i s = |\frac{20 + 72 + 51}{13}| = \frac{143}{13} = 11$

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Let a circle C of radius 5 lie below the x-axis. The line L1=4x+3y+2=0 passes through the centre P of the circle C and intersects the line L2:3x−4y−11=0 at Q. The line L2 touches C at the point Q. Then the distance of P from the line 5x-12y+51=0 is