Question:
The circle touching the coordinate axes with its centre lying on
\[
x-2y-3=0
\]
is:
The circle touching the coordinate axes with its centre lying on
\[
x-2y-3=0
\]
is:
Show Hint
A circle touching both coordinate axes has centre \((\pm r,\pm r)\), because its perpendicular distance from both axes must be equal to its radius.
Updated On: Jun 24, 2026
- \(x^2+y^2-2x+2y+1=0\)
- \(x^2+y^2+2x-2y+1=0\)
- \(x^2+y^2+6x+6y-9=0\)
- \(x^2+y^2-6x-6y+9=0\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the condition of touching both coordinate axes.
If a circle touches both coordinate axes, then the absolute values of the coordinates of its centre are equal to the radius.
So, the centre must be of the form \[ (r,r),\quad (r,-r),\quad (-r,r),\quad \text{or} \quad (-r,-r) \]
Step 2: Check option (1).
The given circle in option (1) is \[ x^2+y^2-2x+2y+1=0 \] Compare with the general equation \[ x^2+y^2+2gx+2fy+c=0 \] Here, \[ 2g=-2,\quad 2f=2,\quad c=1 \] So, \[ g=-1,\quad f=1 \] The centre is \[ (-g,-f)=(1,-1) \] The radius is \[ \sqrt{g^2+f^2-c} \] \[ =\sqrt{(-1)^2+(1)^2-1} \] \[ =\sqrt{1} \] \[ =1 \] Thus, the centre is \[ (1,-1) \] and radius is \[ 1 \] Since the distances of \((1,-1)\) from both coordinate axes are \(1\), the circle touches both coordinate axes.
Step 3: Check whether the centre lies on the given line.
The given line is \[ x-2y-3=0 \] Substitute the centre \[ (1,-1) \] \[ 1-2(-1)-3=0 \] \[ 1+2-3=0 \] \[ 0=0 \] Hence, the centre lies on \[ x-2y-3=0 \]
Step 4: Final conclusion.
Therefore, the required circle is \[ \boxed{x^2+y^2-2x+2y+1=0} \]
If a circle touches both coordinate axes, then the absolute values of the coordinates of its centre are equal to the radius.
So, the centre must be of the form \[ (r,r),\quad (r,-r),\quad (-r,r),\quad \text{or} \quad (-r,-r) \]
Step 2: Check option (1).
The given circle in option (1) is \[ x^2+y^2-2x+2y+1=0 \] Compare with the general equation \[ x^2+y^2+2gx+2fy+c=0 \] Here, \[ 2g=-2,\quad 2f=2,\quad c=1 \] So, \[ g=-1,\quad f=1 \] The centre is \[ (-g,-f)=(1,-1) \] The radius is \[ \sqrt{g^2+f^2-c} \] \[ =\sqrt{(-1)^2+(1)^2-1} \] \[ =\sqrt{1} \] \[ =1 \] Thus, the centre is \[ (1,-1) \] and radius is \[ 1 \] Since the distances of \((1,-1)\) from both coordinate axes are \(1\), the circle touches both coordinate axes.
Step 3: Check whether the centre lies on the given line.
The given line is \[ x-2y-3=0 \] Substitute the centre \[ (1,-1) \] \[ 1-2(-1)-3=0 \] \[ 1+2-3=0 \] \[ 0=0 \] Hence, the centre lies on \[ x-2y-3=0 \]
Step 4: Final conclusion.
Therefore, the required circle is \[ \boxed{x^2+y^2-2x+2y+1=0} \]
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