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

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(C) Given the lines $L_{1}: 4x + 3y + 2 = 0$ and $L_{2}: 3x - 4y - 11 = 0$. The intersection point $Q$ of these two lines is the point of tangency.Sol

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$11$

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(C) Given the lines $L_{1}: 4x + 3y + 2 = 0$ and $L_{2}: 3x - 4y - 11 = 0$. The intersection point $Q$ of these two lines is the point of tangency.Solving the system of equations:$4x + 3y = -2$ (multiply by $4$): $16x + 12y = -8$$3x - 4y = 11$ (multiply by $3$): $9x - 12y = 33$Adding these gives $25x = 25$,so $x = 1$.Substituting $x = 1$ into $4(1) + 3y = -2$,we get $3y = -6$,so $y = -2$. Thus,$Q = (1, -2)$.The line $L_{1}$ is the normal to the circle at $Q$ because it passes through the center $P$. The slope of $L_{2}$ is $m = \frac{3}{4}$. The slope of the normal $L_{1}$ is $-\frac{4}{3}$.The center $P$ lies on $L_{1}$ at a distance of $5$ units from $Q$ along the normal. The unit vector along the normal $L_{1}$ (direction $(3, -4)$) is $(\frac{3}{5}, -\frac{4}{5})$.Since the circle lies below the $x$-axis,the center $P = Q + 5(\frac{3}{5}, -\frac{4}{5}) = (1 + 3, -2 - 4) = (4, -6)$.Now,calculate the distance of $P(4, -6)$ from the line $5x - 12y + 51 = 0$:Distance $= \left| \frac{5(4) - 12(-6) + 51}{\sqrt{5^2 + (-12)^2}} \right| = \left| \frac{20 + 72 + 51}{13} \right| = \frac{143}{13} = 11$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

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