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प्रश्न
One diameter of the circle circumscribing the rectangle ABCD is 4y = x + 7. If the coordinates of A and B are (−3, 4) and (5, 4) respectively, find the equation of the circle.
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उत्तर
Clearly, the centre of the circle lies on the line 4y = x + 7.
The circle passes through A (−3, 4) and B (5, 4).
The slope of the segment joining A and B is zero.
Therefore, the slope of the perpendicular bisector of AB is not defined.
Hence, the perpendicular bisector of AB will be parallel to the y-axis and will pass through \[\left( \frac{- 3 + 5}{2}, \frac{4 + 4}{2} \right) = \left( 1, 4 \right)\]
The equation of the perpendicular bisector is \[x = 1\]
The intersection point of the perpendicular bisector and 4y = x + 7 is \[\left( 1, 2 \right)\]
∴ Centre =\[\left( 1, 2 \right)\]
Radius = \[\sqrt{\left( 5 - 1 \right)^2 + \left( 4 - 2 \right)^2} = \sqrt{20}\]
Hence, the required equation of the circle is \[x^2 + y^2 - 2x - 4y - 15 = 0\]
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