Question:
The equation of the circle passing through the points of intersection of the circles
\[
x^2+y^2-2x-4y+1=0,
\]
\[
x^2+y^2-4x-2y+4=0
\]
and having its centre on the line
\[
x-2y-3=0
\]
is
The equation of the circle passing through the points of intersection of the circles
\[
x^2+y^2-2x-4y+1=0,
\]
\[
x^2+y^2-4x-2y+4=0
\]
and having its centre on the line
\[
x-2y-3=0
\]
is
Show Hint
If a circle passes through the common points of two circles \(S_1=0\) and \(S_2=0\), use
\[
S_1+\lambda(S_2-S_1)=0.
\]
Then apply the extra condition (centre, radius, tangent, etc.) to determine \(\lambda\).
Updated On: Jul 29, 2026
- \[ x^2+y^2-6x+7=0 \]
- \[ x^2+y^2+6x+7=0 \]
- \[ x^2+y^2+6x-7=0 \]
- \[ x^2+y^2-6x-7=0 \]
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
All circles passing through the points of intersection of the circles
\[
S_1=0
\]
and
\[
S_2=0
\]
are represented by
\[
S_1+\lambda S_2=0.
\]
Equivalently,
\[
S_1+\mu(S_2-S_1)=0.
\]
The required parameter is determined using the condition on the centre.
Step 1: Write the family of circles through the common points. Let \[ S_1=x^2+y^2-2x-4y+1, \] \[ S_2=x^2+y^2-4x-2y+4. \] Then \[ S_2-S_1=-2x+2y+3. \] Hence the required family is \[ S=S_1+\lambda(S_2-S_1)=0. \] \[ x^2+y^2+(-2-2\lambda)x+(-4+2\lambda)y+(1+3\lambda)=0. \]
Step 2: Find the centre of this circle. Comparing with \[ x^2+y^2+2gx+2fy+c=0, \] we get \[ g=-1-\lambda, \] \[ f=-2+\lambda. \] Therefore, the centre is \[ (-g,-f) = (1+\lambda,\;2-\lambda). \]
Step 3: Use the condition that the centre lies on \(x-2y-3=0\). Substituting \[ x=1+\lambda, \qquad y=2-\lambda, \] into \[ x-2y-3=0, \] we obtain \[ (1+\lambda)-2(2-\lambda)-3=0. \] \[ 1+\lambda-4+2\lambda-3=0. \] \[ 3\lambda-6=0. \] \[ \lambda=2. \]
Step 4: Substitute \(\lambda=2\) into the family. \[ x^2+y^2+(-2-4)x+(-4+4)y+(1+6)=0. \] \[ x^2+y^2-6x+7=0. \]
Step 5: Write the final answer. \[ \boxed{x^2+y^2-6x+7=0} \]
Step 1: Write the family of circles through the common points. Let \[ S_1=x^2+y^2-2x-4y+1, \] \[ S_2=x^2+y^2-4x-2y+4. \] Then \[ S_2-S_1=-2x+2y+3. \] Hence the required family is \[ S=S_1+\lambda(S_2-S_1)=0. \] \[ x^2+y^2+(-2-2\lambda)x+(-4+2\lambda)y+(1+3\lambda)=0. \]
Step 2: Find the centre of this circle. Comparing with \[ x^2+y^2+2gx+2fy+c=0, \] we get \[ g=-1-\lambda, \] \[ f=-2+\lambda. \] Therefore, the centre is \[ (-g,-f) = (1+\lambda,\;2-\lambda). \]
Step 3: Use the condition that the centre lies on \(x-2y-3=0\). Substituting \[ x=1+\lambda, \qquad y=2-\lambda, \] into \[ x-2y-3=0, \] we obtain \[ (1+\lambda)-2(2-\lambda)-3=0. \] \[ 1+\lambda-4+2\lambda-3=0. \] \[ 3\lambda-6=0. \] \[ \lambda=2. \]
Step 4: Substitute \(\lambda=2\) into the family. \[ x^2+y^2+(-2-4)x+(-4+4)y+(1+6)=0. \] \[ x^2+y^2-6x+7=0. \]
Step 5: Write the final answer. \[ \boxed{x^2+y^2-6x+7=0} \]
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