Let a circle C of radius 5 lie below the x-ax | vedclass

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Question

$4 x+3 y+2=0$

$3 x-4 y-11=0$

$\frac{x}{-25}=\frac{y}{50}=\frac{1}{-25}$

$\frac{x-1}{\cos 	heta}=\frac{y+2}{\sin 	heta}=\pm 5$

$y=-2+5\le

Vedclass · Accepted Answer

$11$

Vedclass · Answer

$4 x+3 y+2=0$

$3 x-4 y-11=0$

$\frac{x}{-25}=\frac{y}{50}=\frac{1}{-25}$

$\frac{x-1}{\cos 	heta}=\frac{y+2}{\sin 	heta}=\pm 5$

$y=-2+5\left(-\frac{4}{5}ight)=-6$

$x=1+5\left(\frac{3}{5}ight)=4$

Req. distance

$\left|\frac{5(4)-12(-6)+51}{13}ight|$

$=\left|\frac{20+72+51}{13}ight|$

$=\frac{143}{13}=11$

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 $P$ 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

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