Let alpha be an integer multiple of 8. If S i | vedclass

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(A) The equation of the circle is $x^2 + y^2 - 4 x - 6 y + 9 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2, f = -3, c = 9$. Cent

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$4$

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(A) The equation of the circle is $x^2 + y^2 - 4 x - 6 y + 9 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2, f = -3, c = 9$. Centre $= (-g, -f) = (2, 3)$ and Radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - 9} = 2$. The line $6 x + 8 y + \alpha = 0$ intersects the circle at two distinct points if the perpendicular distance from the centre $(2, 3)$ to the line is less than the radius. Distance $d = \frac{|6(2) + 8(3) + \alpha|}{\sqrt{6^2 + 8^2}} = \frac{|12 + 24 + \alpha|}{10} = \frac{|36 + \alpha|}{10}$. Condition: $d This implies $-20 Given $\alpha$ is an integer multiple of $8$,let $\alpha = 8k$ for some integer $k$. $-56 The possible integer values for $k$ are $-6, -5, -4, -3$. Thus,there are $4$ possible values for $\alpha$.

Let $\alpha$ be an integer multiple of $8$. If $S$ is the set of all possible values of $\alpha$ such that the line $6 x + 8 y + \alpha = 0$ intersects the circle $x^2 + y^2 - 4 x - 6 y + 9 = 0$ at two distinct points,then the number of elements in $S$ is

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