Question:
The point diametrically opposite to the point \((-1,1)\) on the circle \(x^{2}+y^{2}-x-4 y+1=0\) is
The point diametrically opposite to the point \((-1,1)\) on the circle \(x^{2}+y^{2}-x-4 y+1=0\) is
Show Hint
The centre is the midpoint of the diameter; find the centre and reflect the given point through it.
Updated On: Nov 12, 2026
- \((-2,5)\)
- \((2,-5)\)
- \((0,3)\)
- \((2,3)\)
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The Correct Option is D
Solution and Explanation
For \(x^2+y^2-x-4y+1=0\) the centre is \(\left(\frac{1}{2},2\right)\). The point \((-1,1)\) satisfies the equation: \(1+1+1-4+1=0\).
If the opposite end is \((h,k)\), then \(\frac{-1+h}{2}=\frac{1}{2}\) and \(\frac{1+k}{2}=2\).
So \(h=2\) and \(k=3\); the point is \((2,3)\).
Answer: (4) \((2,3)\)
If the opposite end is \((h,k)\), then \(\frac{-1+h}{2}=\frac{1}{2}\) and \(\frac{1+k}{2}=2\).
So \(h=2\) and \(k=3\); the point is \((2,3)\).
Answer: (4) \((2,3)\)
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