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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. - Mathematics

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Question

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.

Solution

Clearly, the centre of the circle lies on the line 4y = x + 7.
The circle passes through A (−3, 4) and (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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Circle - Standard Equation of a Circle
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Chapter 24: The circle - Exercise 24.1 [Page 22]

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RD Sharma Mathematics [English] Class 11
Chapter 24 The circle
Exercise 24.1 | Q 20 | Page 22

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