Question:
If a circle passes through the points \((2,3)\) and \((4,5)\) and its center lies on the straight line \(y-4x+3 = 0\), then its equation is......
If a circle passes through the points \((2,3)\) and \((4,5)\) and its center lies on the straight line \(y-4x+3 = 0\), then its equation is......
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Centre lies on the perpendicular bisector of the chord and on the given line.
Updated On: Oct 1, 2026
- \(x^2+y^2-4x-10y+25 = 0\)
- \(x^2+y^2-4x-10y-25 = 0\)
- \(x^2+y^2-4x+10y-25 = 0\)
- \(x^2+y^2+25 = 0\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
The centre is equidistant from \((2,3)\) and \((4,5)\), so it lies on the perpendicular bisector of the segment joining them. It also lies on \(y=4x-3\).
Step 2: Key Formula or Approach
Midpoint \((3,4)\), slope of the chord \(=1\), so the bisector is \(x+y=7\).
Step 3: Detailed Explanation
Solve \(y=4x-3\) and \(x+y=7\): \(5x-3=7\), \(x=2\), \(y=5\). Centre \((2,5)\).
Radius squared: distance to \((2,3)\): \(r^2=0+4=4\).
Equation: \((x-2)^2+(y-5)^2=4\), i.e.
\[ x^2+y^2-4x-10y+25=0 \]
Final Answer:
The circle is \(x^2+y^2-4x-10y+25=0\), option (A). \[ \boxed{x^2+y^2-4x-10y+25=0\ \text{(A)}} \]
The centre is equidistant from \((2,3)\) and \((4,5)\), so it lies on the perpendicular bisector of the segment joining them. It also lies on \(y=4x-3\).
Step 2: Key Formula or Approach
Midpoint \((3,4)\), slope of the chord \(=1\), so the bisector is \(x+y=7\).
Step 3: Detailed Explanation
Solve \(y=4x-3\) and \(x+y=7\): \(5x-3=7\), \(x=2\), \(y=5\). Centre \((2,5)\).
Radius squared: distance to \((2,3)\): \(r^2=0+4=4\).
Equation: \((x-2)^2+(y-5)^2=4\), i.e.
\[ x^2+y^2-4x-10y+25=0 \]
Final Answer:
The circle is \(x^2+y^2-4x-10y+25=0\), option (A). \[ \boxed{x^2+y^2-4x-10y+25=0\ \text{(A)}} \]
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