Question:
Let \(\mathrm{A}(1,2)\) and \(\mathrm{B}(3,6)\) be two points on a circle of radius \(\mathrm{R}\), whose one of the diameters is along the line \(2x-y+4=0\). Then \(\mathrm{R}^2\) is equal to :
Let \(\mathrm{A}(1,2)\) and \(\mathrm{B}(3,6)\) be two points on a circle of radius \(\mathrm{R}\), whose one of the diameters is along the line \(2x-y+4=0\). Then \(\mathrm{R}^2\) is equal to :
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The centre lies on the given diameter and on the perpendicular bisector of the chord AB; find it and compute its distance from A.
Updated On: Nov 12, 2026
- 17
- \(\frac{37}{3}\)
- \(\frac{31}{5}\)
- \(\frac{41}{5}\)
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The Correct Option is D
Solution and Explanation
The centre lies on the perpendicular bisector of chord AB and on the diameter \(2x-y+4=0\).
Midpoint of AB is \((2,4)\); slope of AB is \(\frac{6-2}{3-1}=2\), so the perpendicular bisector has slope \(-\frac12\): \(y-4=-\frac12(x-2)\), i.e. \(x+2y=10\).
Solve with \(y=2x+4\): \(x+4x+8=10\Rightarrow x=\frac25\), \(y=\frac{24}{5}\). Centre \(C=\left(\frac25,\frac{24}5\right)\).
\(R^2=CA^2=\left(1-\frac25\right)^2+\left(2-\frac{24}5\right)^2=\frac{9}{25}+\frac{196}{25}=\frac{205}{25}=\frac{41}{5}\).
Answer: (4) \(\frac{41}{5}\)
Midpoint of AB is \((2,4)\); slope of AB is \(\frac{6-2}{3-1}=2\), so the perpendicular bisector has slope \(-\frac12\): \(y-4=-\frac12(x-2)\), i.e. \(x+2y=10\).
Solve with \(y=2x+4\): \(x+4x+8=10\Rightarrow x=\frac25\), \(y=\frac{24}{5}\). Centre \(C=\left(\frac25,\frac{24}5\right)\).
\(R^2=CA^2=\left(1-\frac25\right)^2+\left(2-\frac{24}5\right)^2=\frac{9}{25}+\frac{196}{25}=\frac{205}{25}=\frac{41}{5}\).
Answer: (4) \(\frac{41}{5}\)
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