If the line y - 1 = m(x - 1) intersects the c | vedclass

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(C) The line is given by $y - 1 = m(x - 1)$,which simplifies to $mx - y + (1 - m) = 0$.For a line $Ax + By + C = 0$ to intersect the circle $x^2 + y^2

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Infinite

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(C) The line is given by $y - 1 = m(x - 1)$,which simplifies to $mx - y + (1 - m) = 0$.For a line $Ax + By + C = 0$ to intersect the circle $x^2 + y^2 = a^2$ at two distinct points,the perpendicular distance from the center $(0, 0)$ to the line must be less than the radius $a = 2$.The distance $d = \frac{|m(0) - (0) + (1 - m)|}{\sqrt{m^2 + (-1)^2}} = \frac{|1 - m|}{\sqrt{m^2 + 1}}$.Setting $d Squaring both sides: $(1 - m)^2 $1 - 2m + m^2 $3m^2 + 2m + 3 > 0$.Since the discriminant of $3m^2 + 2m + 3$ is $D = 2^2 - 4(3)(3) = 4 - 36 = -32 Thus,there are infinitely many values of $m$ for which the line intersects the circle at two distinct points.

If the line $y - 1 = m(x - 1)$ intersects the circle $x^2 + y^2 = 4$ at two distinct points,then the number of possible values of $m$ is:

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