If the line 4x - 3y + p = 0 (p + 3 0) touch | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 4x + 6y + 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2, f = 3, c = 4$. Centre

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$2$

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(B) The equation of the circle is $x^2 + y^2 - 4x + 6y + 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2, f = 3, c = 4$. Centre $C = (-g, -f) = (2, -3)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - 4} = 3$. Since the line $4x - 3y + p = 0$ is tangent to the circle,the perpendicular distance from the centre $(2, -3)$ to the line must be equal to the radius $r = 3$. $\frac{|4(2) - 3(-3) + p|}{\sqrt{4^2 + (-3)^2}} = 3$ $\Rightarrow \frac{|8 + 9 + p|}{5} = 3$ $\Rightarrow |17 + p| = 15$. This gives $17 + p = 15 \Rightarrow p = -2$ or $17 + p = -15 \Rightarrow p = -32$. Given $p + 3 > 0$,we have $p > -3$,so $p = -2$. The line is $4x - 3y - 2 = 0$. The point of contact $(h, k)$ lies on the line $4h - 3k - 2 = 0$. The normal at $(h, k)$ passes through the centre $(2, -3)$. The slope of the tangent is $4/3$,so the slope of the normal is $-3/4$. $\frac{k - (-3)}{h - 2} = -\frac{3}{4}$ $\Rightarrow 4k + 12 = -3h + 6$ $\Rightarrow 3h + 4k = -6$. Solving $4h - 3k = 2$ and $3h + 4k = -6$: Multiply first by $4$ and second by $3$: $16h - 12k = 8$ and $9h + 12k = -18$. Adding them: $25h = -10 \Rightarrow h = -2/5$. Substituting $h$: $4(-2/5) - 3k = 2$ $\Rightarrow -8/5 - 2 = 3k$ $\Rightarrow 3k = -18/5$ $\Rightarrow k = -6/5$. Thus,$h - 2k = -2/5 - 2(-6/5) = -2/5 + 12/5 = 10/5 = 2$.

If the line $4x - 3y + p = 0$ $(p + 3 > 0)$ touches the circle $x^2 + y^2 - 4x + 6y + 4 = 0$ at the point $(h, k)$,then $h - 2k = . . . . . .$

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