If the straight line 3x + 4y = k touches the | vedclass

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(C) Given the circle equation: $x^2 + y^2 = 16x$ Rearranging the terms: $x^2 - 16x + y^2 = 0$ Completing the square: $(x - 8)^2 + y^2 = 64$ Thus,the c

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$-16, 64$

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(C) Given the circle equation: $x^2 + y^2 = 16x$ Rearranging the terms: $x^2 - 16x + y^2 = 0$ Completing the square: $(x - 8)^2 + y^2 = 64$ Thus,the center is $(8, 0)$ and the radius $r = 8$. Since the line $3x + 4y - k = 0$ touches the circle,the perpendicular distance from the center $(8, 0)$ to the line must equal the radius $r = 8$. Using the distance formula: $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$ $8 = \frac{|3(8) + 4(0) - k|}{\sqrt{3^2 + 4^2}}$ $8 = \frac{|24 - k|}{\sqrt{25}}$ $8 = \frac{|24 - k|}{5}$ $|24 - k| = 40$ Case $1$: $24 - k = 40 \Rightarrow k = -16$ Case $2$: $24 - k = -40 \Rightarrow k = 64$ Therefore,the values of $k$ are $-16$ and $64$.

If the straight line $3x + 4y = k$ touches the circle $x^2 + y^2 = 16x$,then the value of $k$ is

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