Question

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 _____

Answer 1

$L_1:4x+3y+2=0$
$L_2:3x−4y−11=0$
Since circle C touches the line L2 at Q intersection point Q of L1 and L2, is (1, –2)
$∵$ P lies of L1
$∴$ $P(x,\frac{−1}{3}(2+4x))$
Now,
$PQ=5⇒$ $(x−1)^2+(\frac{4x+2}{3}−2)^2=25$
⇒($$x−1)^2[1+\frac{16}{9}]=25
$⇒ (x – 1)^2 = 9$
$⇒ x = 4, –2$
$∵$ The circle lies below the x-axis.
$y = –6$
$P(4, –6)$
Now distance of P from $5x – 12y \+ 51 = 0$
$=$ $\left| \frac{20 + 72 + 51}{13} \right| = \frac{143}{13} = 11$