Let y=x+2,4y=3x+6,and 3y=4x+1 be three tangen | vedclass

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(A) The given lines are $L_1: x-y+2=0$,$L_2: 3x-4y+6=0$,and $L_3: 4x-3y+1=0$.The center $(h, k)$ of the circle is equidistant from these three tangent

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$5$

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(A) The given lines are $L_1: x-y+2=0$,$L_2: 3x-4y+6=0$,and $L_3: 4x-3y+1=0$.The center $(h, k)$ of the circle is equidistant from these three tangent lines,so it must lie on the angle bisectors of the lines.The angle bisectors of $L_2$ and $L_3$ are given by $\frac{3x-4y+6}{5} = \pm \frac{4x-3y+1}{5}$.Case $1$: $3x-4y+6 = 4x-3y+1 \Rightarrow x+y=5$.Case $2$: $3x-4y+6 = -(4x-3y+1)$ $\Rightarrow 7x-7y+7=0$ $\Rightarrow x-y+1=0$.Since the center $(h, k)$ must also be equidistant from $L_1$ and $L_2$,we check the intersection of the bisectors with the locus of points equidistant from $L_1$ and $L_2$.For the circle to be tangent to all three lines,the center $(h, k)$ is the incenter of the triangle formed by the lines. Solving the system,we find the center $(h, k)$ lies on the line $x+y=5$.Thus,$h+k=5$.

Let $y=x+2$,$4y=3x+6$,and $3y=4x+1$ be three tangent lines to the circle $(x-h)^2+(y-k)^2=r^2$. Then $h+k$ is equal to:

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