The straight line (x - 2) + (y + 3) = 0 cuts | vedclass

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(C) The equation of the circle is $(x - 2)^2 + (y - 3)^2 = 11$. The centre is $(2, 3)$ and the radius $r = \sqrt{11}$.The equation of the line is $x +

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(C) The equation of the circle is $(x - 2)^2 + (y - 3)^2 = 11$. The centre is $(2, 3)$ and the radius $r = \sqrt{11}$.The equation of the line is $x + y - 1 = 0$.The perpendicular distance $d$ from the centre $(2, 3)$ to the line $x + y - 1 = 0$ is given by $d = \frac{|2 + 3 - 1|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} = \sqrt{8}$.Since $d = \sqrt{8}$ and $r = \sqrt{11}$,we have $d Because the perpendicular distance from the centre to the line is less than the radius,the line intersects the circle at two distinct points.

The straight line $(x - 2) + (y + 3) = 0$ cuts the circle $(x - 2)^2 + (y - 3)^2 = 11$ at

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