If the line x-2y=m (m in mathbb{Z}) intersect | vedclass

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(B) The equation of the circle is $x^2 - 2x + y^2 - 4y = 0$. Completing the square,we get $(x-1)^2 + (y-2)^2 = 5$. The centre of the circle is $(1, 2)

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$9$

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(B) The equation of the circle is $x^2 - 2x + y^2 - 4y = 0$. Completing the square,we get $(x-1)^2 + (y-2)^2 = 5$. The centre of the circle is $(1, 2)$ and the radius $r = \sqrt{5}$. For the line $x - 2y - m = 0$ to intersect the circle at two distinct points,the perpendicular distance from the centre $(1, 2)$ to the line must be less than the radius $r$. Distance $d = \frac{|1 - 2(2) - m|}{\sqrt{1^2 + (-2)^2}} = \frac{|1 - 4 - m|}{\sqrt{5}} = \frac{|-3 - m|}{\sqrt{5}} = \frac{|m + 3|}{\sqrt{5}}$. Setting $d $|m + 3| $-5 $-8 Since $m \in \mathbb{Z}$,the possible values for $m$ are $\{-7, -6, -5, -4, -3, -2, -1, 0, 1\}$. The total number of such values is $9$.

If the line $x-2y=m$ $(m \in \mathbb{Z})$ intersects the circle $x^2+y^2=2x+4y$ at two distinct points,then the number of possible values of $m$ are

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