Question:
If a circle of radius \(3\) passes through the point \((7,3)\) and has its centre on the line
\[
x-y-1=0,
\]
then its equation among the following is:
If a circle of radius \(3\) passes through the point \((7,3)\) and has its centre on the line
\[
x-y-1=0,
\]
then its equation among the following is:
Show Hint
For \(x^2+y^2+2gx+2fy+c=0\), the centre is \((-g,-f)\) and the radius is \(\sqrt{g^2+f^2-c}\).
Updated On: Jun 18, 2026
- \(x^2+y^2+14x-12y+76=0\)
- \(x^2+y^2+14x-12y+76=0\)
- \(x^2+y^2+8x-6y+16=0\)
- \(x^2+y^2-14x-12y+76=0\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Let the centre of the circle be \((h,k)\).
Since the centre lies on \[ x-y-1=0, \] we have \[ h-k-1=0. \] Therefore, \[ h-k=1. \]
Step 2: Use the radius condition.
The circle has radius \(3\) and passes through \((7,3)\).
So the distance between \((h,k)\) and \((7,3)\) is \(3\).
Hence, \[ (h-7)^2+(k-3)^2=9. \]
Step 3: Identify centre from the given option.
For a circle \[ x^2+y^2+2gx+2fy+c=0, \] the centre is \[ (-g,-f). \] For option (4), \[ x^2+y^2-14x-12y+76=0. \] Here, \[ 2g=-14,\qquad 2f=-12. \] So, \[ g=-7,\qquad f=-6. \] Hence the centre is \[ (-g,-f)=(7,6). \]
Step 4: Verify the centre and radius.
The centre \((7,6)\) lies on \[ x-y-1=0 \] because \[ 7-6-1=0. \] Also, distance from \((7,6)\) to \((7,3)\) is \[ \sqrt{(7-7)^2+(6-3)^2}=3. \] So the radius is \(3\).
Step 5: Final conclusion.
Therefore, the required circle is \[ \boxed{x^2+y^2-14x-12y+76=0} \]
Since the centre lies on \[ x-y-1=0, \] we have \[ h-k-1=0. \] Therefore, \[ h-k=1. \]
Step 2: Use the radius condition.
The circle has radius \(3\) and passes through \((7,3)\).
So the distance between \((h,k)\) and \((7,3)\) is \(3\).
Hence, \[ (h-7)^2+(k-3)^2=9. \]
Step 3: Identify centre from the given option.
For a circle \[ x^2+y^2+2gx+2fy+c=0, \] the centre is \[ (-g,-f). \] For option (4), \[ x^2+y^2-14x-12y+76=0. \] Here, \[ 2g=-14,\qquad 2f=-12. \] So, \[ g=-7,\qquad f=-6. \] Hence the centre is \[ (-g,-f)=(7,6). \]
Step 4: Verify the centre and radius.
The centre \((7,6)\) lies on \[ x-y-1=0 \] because \[ 7-6-1=0. \] Also, distance from \((7,6)\) to \((7,3)\) is \[ \sqrt{(7-7)^2+(6-3)^2}=3. \] So the radius is \(3\).
Step 5: Final conclusion.
Therefore, the required circle is \[ \boxed{x^2+y^2-14x-12y+76=0} \]
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