Question:
The equation of the circle touching the lines \(|x-2| + |y-3| = 4\) is?
The equation of the circle touching the lines \(|x-2| + |y-3| = 4\) is?
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Use the condition of tangency of a circle to given lines and find the circle equation satisfying those conditions.
Updated On: Jun 6, 2025
- \(x^2 + y^2 - 6x - 4y + 5 = 0\)
- \(x^2 + y^2 - 4x - 6y + 5 = 0\)
- \(x^2 + y^2 - x - 2y - 5 = 0\)
- \(x^2 + y^2 - 2x - y - 5 = 0\)
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The Correct Option is B
Solution and Explanation
Given the circle touches the lines \(|x-2| + |y-3| = 4\), which represent boundary lines of a square or diamond shaped region.
The circle must be tangent to these lines, so its radius equals the distance from center to each line.
Center of circle \((h,k)\) can be assumed, and from tangency conditions the circle equation is found to be
\[
x^2 + y^2 - 4x - 6y + 5 = 0.
\]
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