Let a circle C of radius 5 lie below the x -axis. The line L 1 = 4 x + 3 y + 2 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 L 2 touches C at the point Q . Then the distance of P from the line 5 x - 12 y + 51 = 0 is

Question

Aicharya · Accepted Answer

Plotting the diagram of given data, we get

Finding intersection point of $4 x + 3 y + 2 = 0$ and $3 x - 4 y - 11 = 0$ we get $(1, - 2)$

Now finding the point $P$ by parametric we get,

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

$y = - 2 + 5 (- \frac{4}{5}) = - 6$

$x = 1 + 5 (\frac{3}{5}) = 4$

Now finding distance of point $P$ from line $5 x - 12 y + 51 = 0$ $= |\frac{5 (4) - 12 (- 6) + 51}{13}|$

$= |\frac{20 + 72 + 51}{13}|$

$= \frac{143}{13} = 11$

Let a circle C of radius 5 lie below the x -axis. The line L 1 = 4 x + 3 y + 2 passes through the centre P…

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