The number of integral values of k for which | vedclass

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(A) The given circle is $x^{2} + y^{2} - 2x - 4y + 4 = 0$.Completing the square,we get $(x - 1)^{2} + (y - 2)^{2} = 1$.Thus,the centre is $(1, 2)$ and

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$9$

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(A) The given circle is $x^{2} + y^{2} - 2x - 4y + 4 = 0$.Completing the square,we get $(x - 1)^{2} + (y - 2)^{2} = 1$.Thus,the centre is $(1, 2)$ and the radius $r = 1$.For the line $3x + 4y - k = 0$ to intersect the circle at two distinct points,the perpendicular distance from the centre to the line must be less than the radius.Distance $d = \frac{|3(1) + 4(2) - k|}{\sqrt{3^{2} + 4^{2}}} $\frac{|11 - k|}{5} $|11 - k| $-5 $-16 $6 The integral values of $k$ are $7, 8, 9, 10, 11, 12, 13, 14, 15$.The total number of such values is $9$.

The number of integral values of $k$ for which the line $3x + 4y = k$ intersects the circle $x^{2} + y^{2} - 2x - 4y + 4 = 0$ at two distinct points is

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