Question:
The equation of a circle which passes through the points \((2,3)\) and \((4,5)\) and whose center lies on the straight line \(y-4x+3 = 0\) is
The equation of a circle which passes through the points \((2,3)\) and \((4,5)\) and whose center lies on the straight line \(y-4x+3 = 0\) is
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The centre is equidistant from both points and lies on $y=4x-3$.
Updated On: Oct 1, 2026
- \(x^2+y^2-4x-10y+25 = 0\)
- \(x^2+y^2-4x-10y-25 = 0\)
- \(x^2+y^2-4x+10y-25 = 0\)
- \(x^2+y^2+4x-10y+25 = 0\)
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The Correct Option is A
Solution and Explanation
Step 1: Set up
Let the centre be \((h,k)\). It lies on \(y-4x+3=0\), so \(k=4h-3\).
Step 2: Equal distances
The distance to \((2,3)\) equals the distance to \((4,5)\):
\[ (h-2)^2+(k-3)^2=(h-4)^2+(k-5)^2 \]
\[ -4h-6k+13=-8h-10k+41\;\Rightarrow\;h+k=7 \]
Step 3: Solve
Put \(k=4h-3\): \(5h-3=7\), so \(h=2\) and \(k=5\). Centre \((2,5)\).
Radius\(^2=(2-2)^2+(3-5)^2=4\).
Step 4: Equation
\[ (x-2)^2+(y-5)^2=4\;\Rightarrow\;x^2+y^2-4x-10y+25=0 \]
This is option (A).
Final Answer:
The circle is \(x^2+y^2-4x-10y+25=0\), option (A). \[ \boxed{\text{(A)}} \]
Let the centre be \((h,k)\). It lies on \(y-4x+3=0\), so \(k=4h-3\).
Step 2: Equal distances
The distance to \((2,3)\) equals the distance to \((4,5)\):
\[ (h-2)^2+(k-3)^2=(h-4)^2+(k-5)^2 \]
\[ -4h-6k+13=-8h-10k+41\;\Rightarrow\;h+k=7 \]
Step 3: Solve
Put \(k=4h-3\): \(5h-3=7\), so \(h=2\) and \(k=5\). Centre \((2,5)\).
Radius\(^2=(2-2)^2+(3-5)^2=4\).
Step 4: Equation
\[ (x-2)^2+(y-5)^2=4\;\Rightarrow\;x^2+y^2-4x-10y+25=0 \]
This is option (A).
Final Answer:
The circle is \(x^2+y^2-4x-10y+25=0\), option (A). \[ \boxed{\text{(A)}} \]
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