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A Circle Whose Centre is the Point of Intersection of the Lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 Passes Through the Origin. Find Its Equation.

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Question

A circle whose centre is the point of intersection of the lines 2x − 3y + 4 = 0 and 3x + 4y− 5 = 0 passes through the origin. Find its equation.

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Solution

Let the required equation of the circle be

\[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]
The point of intersection of the lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0  is
\[\left( \frac{- 1}{17}, \frac{22}{17} \right)\]
∴ Centre = \[\left( \frac{- 1}{17}, \frac{22}{17} \right)\]
Also, the circle passes through the origin.
∴ \[a^2 = \left( \frac{1}{17} \right)^2 + \left( \frac{22}{17} \right)^2 = \frac{485}{289}\]
Hence, the required equation of the circle is
\[\left( x + \frac{1}{17} \right)^2 + \left( y - \frac{22}{17} \right)^2 = \frac{485}{289}\]
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Advanced Concept of Circle - Standard Equation of a Circle
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Chapter 24: The circle - Exercise 24.1 [Page 21]

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R.D. Sharma Mathematics [English] Class 11
Chapter 24 The circle
Exercise 24.1 | Q 10 | Page 21

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