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:

• A. \(x^2+y^2-6x-2(4+\sqrt3)y+(24+8\sqrt3)=0\)
• B. \(x^2+y^2-6x+2(4+\sqrt3)y+(24+8\sqrt3)=0\)
• C. \(x^2+y^2+6x-2(4+\sqrt3)y+(24+8\sqrt3)=0\)
• D. \(x^2+y^2+6x+2(4+\sqrt3)y+(24+8\sqrt3)=0\)

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.

Answer 1

The Correct Option is A

Step 1: Determine the centre and radius of the original circle.
From x² + y² - 4x - 8y + 16 = 0, comparing with x² + y² + 2gx + 2fy + c = 0 gives 2g = -4 → g = -2, 2f = -8 → f = -4, and c = 16. The centre is (2, 4) and radius r = √(g² + f² - c) = √(4 + 16 - 16) = 2.

Step 2: Find the direction of the tangent at the point of contact.
The point of contact is (2 + √3, 3). The radius vector from the centre to this point is (√3, -1). A tangent direction perpendicular to this is (1, √3), whose magnitude is √(1 + 3) = 2. The unit tangent vector is (1/2, √3/2).

Step 3: Displace the centre along the tangent by the rolling distance.
Rolling upward by 2 units means displacement = 2 × (1/2, √3/2) = (1, √3). The new centre becomes (2, 4) + (1, √3) = (3, 4 + √3). The radius remains 2.

Step 4: Write the equation of the new circle.
(x - 3)² + [y - (4 + √3)]² = 4. Expanding: x² - 6x + 9 + y² - 2(4 + √3)y + (16 + 8√3 + 3) = 4 → x² + y² - 6x - 2(4 + √3)y + (24 + 8√3) = 0.

Step 5: Final conclusion.
The equation of the circle after rolling is x² + y² - 6x - 2(4 + √3)y + (24 + 8√3) = 0.