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Find the Equation of the Circle Which Passes Through the Origin and Cuts off Intercepts a and B Respectively from X and Y - Axes.

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Question

Find the equation of the circle which passes through the origin and cuts off intercepts aand b respectively from x and - axes.

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Solution

Case I:
If the required circle passes through the origin and (ab), then the end points of the diameter of the circle will be (0, 0) and (ab).
∴ Required equation of circle: \[\left( x - 0 \right)\left( x - a \right) + \left( y - 0 \right)\left( y - b \right)\] or

\[x^2 + y^2 - ax - by = 0\]
Case II:
If the required circle passes through the origin and (−a, −b), then the end points of the diameter of the circle will be (0, 0) and (−a, −b).
∴ Required equation of circle: \[\left( x - 0 \right)\left( x + a \right) + \left( y - 0 \right)\left( y + b \right)\] or
\[x^2 + y^2 + ax + by = 0\]
Hence, the equation of the required circle is
\[x^2 + y^2 \pm ax \pm by = 0\]
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Advanced Concept of Circle - Standard Equation of a Circle
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Chapter 24: The circle - Exercise 24.3 [Page 37]

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R.D. Sharma Mathematics [English] Class 11
Chapter 24 The circle
Exercise 24.3 | Q 6 | Page 37

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