For what value of m does the line 3x + 4y = m | vedclass

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(C) The given circle is $x^2 + y^2 - 2x - 8 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = 0, c = -8$. The center is $(-g, -

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$18, -12$

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(C) The given circle is $x^2 + y^2 - 2x - 8 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = 0, c = -8$. The center is $(-g, -f) = (1, 0)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 0^2 - (-8)} = \sqrt{1 + 8} = \sqrt{9} = 3$. $A$ line $Ax + By + C = 0$ touches the circle if the perpendicular distance from the center to the line equals the radius. The line is $3x + 4y - m = 0$. Distance $d = \frac{|3(1) + 4(0) - m|}{\sqrt{3^2 + 4^2}} = 3$. $\frac{|3 - m|}{\sqrt{9 + 16}} = 3$. $|3 - m| = 3 	imes 5 = 15$. $3 - m = 15$ or $3 - m = -15$. $m = 3 - 15 = -12$ or $m = 3 + 15 = 18$. Thus,$m = 18, -12$.

For what value of $m$ does the line $3x + 4y = m$ touch the circle $x^2 + y^2 - 2x - 8 = 0$?

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