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Prove that if (x, y) be a point on the circle with the line joining (0, 1) and (2, 3) as diameter, then x2 + y2 – 2x – 4y + 3 = 0. Find the coordinates of the points where this circle cuts they axis. |
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Answer» As (0, 1), (2, 3) are the points joining the diameter of the circle, the center will be \(=(\frac{0+2}{2},\frac{1+3}{2})=(1,2)\) Radius = \(\sqrt{(1-0)^2+(2-1)^2}\) \(\Rightarrow\) r = \(\sqrt{1^2+1^2}=\sqrt{2}\) (x, y) lies on the cirlce (x - 1)2 + (y - 2)2 = \(\sqrt{2}^2\) x2 - 2x + 1 + y2 - 4y + 4 = 2 x2 + y2 - 2x - 4y + 5 - 2 = 0 x2 + y2 - 2x - 4y + 3 = 0 Let (x, 0) be the coordinate which intersect the x axis then, x2 – 2x + 3 = 0 \(x=\frac{2\pm\sqrt{4-12}}{2}=\frac{2\pm\sqrt{-8}}{2}\) This circle can not be intersect the x axis. Let (0, y) be the coordinate which intersect the y axis then, y – 4y + 3 = 0, (y – 3)(y – 1) = 0 Coordinates are (0, 1) and (0, 3) |
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