Question:
The equation of circle with centre \((2,-3)\) and touching \(x\)-axis is
The equation of circle with centre \((2,-3)\) and touching \(x\)-axis is
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If a circle touches the \(x\)-axis, then its radius is equal to the absolute value of the \(y\)-coordinate of its centre.
\[
r=|k|
\]
Updated On: Jun 26, 2026
- \(x^2+y^2-4x-6y+4=0\)
- \(x^2+y^2-4x-6y-8=0\)
- \(x^2+y^2-4x+6y+4=0\)
- \(x^2+y^2+4x-6y+8=0\)
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The Correct Option is C
Solution and Explanation
Step 1: Identify the centre of the circle.
The centre of the circle is given as \[ (2,-3) \] So, \[ h=2,\quad k=-3 \]
Step 2: Find the radius of the circle.
The circle touches the \(x\)-axis.
The radius is equal to the perpendicular distance of the centre from the \(x\)-axis.
For a point \((h,k)\), distance from the \(x\)-axis is \[ |k| \] Here, \[ k=-3 \] Therefore, \[ r=|-3|=3 \]
Step 3: Use the standard equation of circle.
The equation of a circle with centre \((h,k)\) and radius \(r\) is \[ (x-h)^2+(y-k)^2=r^2 \] Substituting \(h=2\), \(k=-3\), and \(r=3\), we get \[ (x-2)^2+(y+3)^2=3^2 \] \[ (x-2)^2+(y+3)^2=9 \]
Step 4: Expand the equation.
Now, \[ (x-2)^2=x^2-4x+4 \] and \[ (y+3)^2=y^2+6y+9 \] So, \[ x^2-4x+4+y^2+6y+9=9 \] \[ x^2+y^2-4x+6y+13=9 \] \[ x^2+y^2-4x+6y+4=0 \]
Step 5: Final conclusion.
Therefore, the required equation of the circle is \[ \boxed{x^2+y^2-4x+6y+4=0} \]
The centre of the circle is given as \[ (2,-3) \] So, \[ h=2,\quad k=-3 \]
Step 2: Find the radius of the circle.
The circle touches the \(x\)-axis.
The radius is equal to the perpendicular distance of the centre from the \(x\)-axis.
For a point \((h,k)\), distance from the \(x\)-axis is \[ |k| \] Here, \[ k=-3 \] Therefore, \[ r=|-3|=3 \]
Step 3: Use the standard equation of circle.
The equation of a circle with centre \((h,k)\) and radius \(r\) is \[ (x-h)^2+(y-k)^2=r^2 \] Substituting \(h=2\), \(k=-3\), and \(r=3\), we get \[ (x-2)^2+(y+3)^2=3^2 \] \[ (x-2)^2+(y+3)^2=9 \]
Step 4: Expand the equation.
Now, \[ (x-2)^2=x^2-4x+4 \] and \[ (y+3)^2=y^2+6y+9 \] So, \[ x^2-4x+4+y^2+6y+9=9 \] \[ x^2+y^2-4x+6y+13=9 \] \[ x^2+y^2-4x+6y+4=0 \]
Step 5: Final conclusion.
Therefore, the required equation of the circle is \[ \boxed{x^2+y^2-4x+6y+4=0} \]
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