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Write the coordinate of the centre of a circle passing through the points (9, 3), (7,-1) (1,-1). Find the radius of the circle. |
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Answer» Assume that (x, y) is centre of the circle Distance between (9,3) and (x,y) = \(\sqrt{(x-9)^2+(y-3)^2}\) Distance between (7,-1) and \((x,y)\) = \(\sqrt{(x+7)^2+(y-1)^2}\) They are equal \(\therefore \sqrt{(x-7)^2+(y-1)^2}\) = \(\sqrt{(x-1)^2+(y-1)^2}\) \((x-7)^2=(x-1)^2\) \(x^2-14x+49=x^2-2x+1\) \(\sqrt{(x-9)^2+(y-3)^2}\) = \(\sqrt{(x-1)^2+(y-1)^2}\) when x = 4 (4 - 9)2 + y2 - 6y + 9 = (4 - 7)2 + y2 + 2y + 1 25 + y2 - 6y + 9 = 9 + y2 + 2y + 1 24 = 8y \(y=\frac{24}{8}=3\) The coordinate of the centre of the circle is (4, 3) ∴ Radius = Distance between (4, 3) and (1,-1) = \(\sqrt{(4-1)^2+(3+1)^2}\) = \(\sqrt{3^2+4^2}=5\) |
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