Question:
The circle
\[
x^2+y^2-4x-8y+16=0
\]
rolls up along the tangent drawn to it at \((2+\sqrt3,3)\) by \(2\) units. The equation of the circle in the new position is:
The circle
\[
x^2+y^2-4x-8y+16=0
\]
rolls up along the tangent drawn to it at \((2+\sqrt3,3)\) by \(2\) units. The equation of the circle in the new position is:
Show Hint
When a circle rolls along a tangent, its centre shifts parallel to the tangent direction by the given distance, while the radius remains unchanged.
Updated On: Jun 18, 2026
- \(x^2+y^2-6x-2(4+\sqrt3)y+(24+8\sqrt3)=0\)
- \(x^2+y^2-6x+2(4+\sqrt3)y+(24+8\sqrt3)=0\)
- \(x^2+y^2+6x-2(4+\sqrt3)y+(24+8\sqrt3)=0\)
- \(x^2+y^2+6x+2(4+\sqrt3)y+(24+8\sqrt3)=0\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Find the centre and radius of the given circle.
Given, \[ x^2+y^2-4x-8y+16=0. \] Comparing with \[ x^2+y^2+2gx+2fy+c=0, \] we get \[ 2g=-4,\quad 2f=-8,\quad c=16. \] So, \[ g=-2,\quad f=-4. \] Hence, the centre is \[ (2,4). \] The radius is \[ r=\sqrt{g^2+f^2-c}. \] \[ r=\sqrt{4+16-16}=2. \]
Step 2: Find the tangent direction.
The point of contact is \[ (2+\sqrt3,3). \] The radius vector from centre \((2,4)\) to this point is \[ (\sqrt3,-1). \] A tangent direction perpendicular to \((\sqrt3,-1)\) is \[ (1,\sqrt3). \] Its magnitude is \[ \sqrt{1^2+(\sqrt3)^2}=2. \] So the unit direction vector is \[ \left(\frac12,\frac{\sqrt3}{2}\right). \]
Step 3: Shift the centre by \(2\) units along the tangent.
Since the circle rolls up by \(2\) units, displacement is \[ 2\left(\frac12,\frac{\sqrt3}{2}\right)=(1,\sqrt3). \] Therefore, the new centre is \[ (2,4)+(1,\sqrt3)=(3,4+\sqrt3). \] The radius remains \[ 2. \]
Step 4: Write the new circle equation.
The new circle is \[ (x-3)^2+(y-(4+\sqrt3))^2=4. \] Expanding, \[ x^2-6x+9+y^2-2(4+\sqrt3)y+(4+\sqrt3)^2=4. \] Now, \[ (4+\sqrt3)^2=16+8\sqrt3+3=19+8\sqrt3. \] So, \[ x^2+y^2-6x-2(4+\sqrt3)y+9+19+8\sqrt3-4=0. \] \[ x^2+y^2-6x-2(4+\sqrt3)y+(24+8\sqrt3)=0. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{x^2+y^2-6x-2(4+\sqrt3)y+(24+8\sqrt3)=0} \]
Given, \[ x^2+y^2-4x-8y+16=0. \] Comparing with \[ x^2+y^2+2gx+2fy+c=0, \] we get \[ 2g=-4,\quad 2f=-8,\quad c=16. \] So, \[ g=-2,\quad f=-4. \] Hence, the centre is \[ (2,4). \] The radius is \[ r=\sqrt{g^2+f^2-c}. \] \[ r=\sqrt{4+16-16}=2. \]
Step 2: Find the tangent direction.
The point of contact is \[ (2+\sqrt3,3). \] The radius vector from centre \((2,4)\) to this point is \[ (\sqrt3,-1). \] A tangent direction perpendicular to \((\sqrt3,-1)\) is \[ (1,\sqrt3). \] Its magnitude is \[ \sqrt{1^2+(\sqrt3)^2}=2. \] So the unit direction vector is \[ \left(\frac12,\frac{\sqrt3}{2}\right). \]
Step 3: Shift the centre by \(2\) units along the tangent.
Since the circle rolls up by \(2\) units, displacement is \[ 2\left(\frac12,\frac{\sqrt3}{2}\right)=(1,\sqrt3). \] Therefore, the new centre is \[ (2,4)+(1,\sqrt3)=(3,4+\sqrt3). \] The radius remains \[ 2. \]
Step 4: Write the new circle equation.
The new circle is \[ (x-3)^2+(y-(4+\sqrt3))^2=4. \] Expanding, \[ x^2-6x+9+y^2-2(4+\sqrt3)y+(4+\sqrt3)^2=4. \] Now, \[ (4+\sqrt3)^2=16+8\sqrt3+3=19+8\sqrt3. \] So, \[ x^2+y^2-6x-2(4+\sqrt3)y+9+19+8\sqrt3-4=0. \] \[ x^2+y^2-6x-2(4+\sqrt3)y+(24+8\sqrt3)=0. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{x^2+y^2-6x-2(4+\sqrt3)y+(24+8\sqrt3)=0} \]
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