Problem 31 - A Complete Resource Book in Mathematics for JEE Main

Question

The circle \(x^{2}+y^{2}-4 x-8 y+16=0\) rolls up the tangent to it at \((2+\sqrt{3}, 3)\) by 2 units, assuming the \(x\)-axis as horizontal, the equation of the circle in the new position is (A) \(x^{2}+y^{2}-6 x-2(4+\sqrt{3}) y+24+8 \sqrt{3}=0\) (B) \(x^{2}+y^{2}+6 x-2(4+\sqrt{3}) y+24+8 \sqrt{3}=0\) (C) \(x^{2}+y^{2}-6 x+2(4+\sqrt{3}) y+24+8 \sqrt{3}=0\) (D) none of these

Step-by-Step Solution

Answer

(A) is the correct answer.

Step 1: Find the center and radius of the original circle

The equation of the circle is given as \(x^{2}+y^{2}-4x-8y+16=0\). To find the center and radius, we first complete the square for both \(x\) and \(y\).Rewrite the equation:\[(x^2 - 4x) + (y^2 - 8y) = -16\]Complete the square:\[(x^2 - 4x + 4) + (y^2 - 8y + 16) = -16 + 4 + 16\]\[(x-2)^2 + (y-4)^2 = 4\]The circle is centered at \((2, 4)\) with a radius of 2.

Step 2: Find the slope of the tangent at the given point

The derivative \((dy/dx)\) of the circle's equation, evaluated at a point \((x, y)\), gives the slope of the tangent line at that point.The implicit differentiation of \(x^2 + y^2 - 4x - 8y + 16 = 0\) gives:\[2x + 2y\frac{dy}{dx} - 4 - 8\frac{dy}{dx} = 0\]\[2(x - 2) + 2y\frac{dy}{dx} = 0\]\[\frac{dy}{dx} = \frac{2x - 4}{8y - 16}\]Substitute \((2+\sqrt{3}, 3)\) into \(\frac{dy}{dx}\) to find the slope:\[m = \frac{2(2+\sqrt{3}) - 4}{8(3) - 16} = \frac{2\sqrt{3}}{8} = \frac{\sqrt{3}}{4}\]

Step 3: Determine the direction vector of the tangent

The direction vector of the tangent line is perpendicular to the radius from the center. We know it has the same direction as the slope \(\frac{\sqrt{3}}{4}\).The direction vector can be taken as \((1, \sqrt{3})\) since it's in the direction where \(1\) unit right corresponds to \(\sqrt{3}\) units up.

Step 4: Calculate new center after moving the circle

The circle rolls upwards along this tangent by 2 units. The correct movement in the \((1, \sqrt{3})\) direction to ensure a total movement of 2 units is calculated using unit vector scaling.Calculate the unit vector:\[\text{Magnitude} = \sqrt{1^2 + (\sqrt{3})^2} = 2\]Therefore, the unit vector is \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\).Move the center \((2, 4)\) by 2 units:\[(2, 4) + 2 \times \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = (2, 4) + (1, \sqrt{3}) = (3, 4+\sqrt{3})\]The new center of the circle is \((3, 4+\sqrt{3})\).

Step 5: Write equation of the circle in new position

The equation of a circle with center \((h, k)\) and radius \(r\) is given by:\[(x-h)^2 + (y-k)^2 = r^2\]Substitute \(h = 3\), \(k = 4+\sqrt{3}\), and \(r = 2\):\[(x - 3)^2 + (y - (4+\sqrt{3}))^2 = 4\]Expand:\[x^2 - 6x + 9 + y^2 - 2(4+\sqrt{3})y + (4+\sqrt{3})^2 = 4\]Calculate \((4+\sqrt{3})^2 = 16 + 8\sqrt{3} + 3 = 19 + 8\sqrt{3}\):Combine terms:\[x^2 + y^2 - 6x - 2(4+\sqrt{3})y + 24 + 8\sqrt{3} = 0\]

Step 6: Match with the given options

The derived equation is:\[x^2 + y^2 - 6x - 2(4+\sqrt{3})y + 24 + 8\sqrt{3} = 0\]This matches option (A) in the list provided:\[x^2 + y^2 - 6x - 2(4+\sqrt{3})y + 24 + 8\sqrt{3} = 0\]

Key Concepts

Circle Equations

Circle equations are a fundamental aspect of coordinate geometry. They define all points at a constant distance, known as the radius, from a fixed point called the center. The most standard form of a circle equation is \[ (x-h)^2 + (y-k)^2 = r^2 \] where

• \((h,k)\) denotes the center of the circle
• \(r\) is the radius of the circle.

In the original exercise, the circle given by \[ x^2 + y^2 - 4x - 8y + 16 = 0 \] is rewritten to identify the center and radius through completing the square. This is a valuable technique used to transform a general circle equation into its standard form by rearranging and grouping like terms. Completing the square reveals the center at \((2, 4)\) and a radius of 2. This transformation facilitates understanding and problem-solving, especially when dealing with concepts like tangents or shifting coordinates.

Tangent to a Circle

A tangent to a circle is a straight line that touches the circumference at exactly one point. At the point of tangency, this line is perpendicular to the radius of the circle. The slope of the tangent is a significant aspect in solving problems involving tangencies. In this context, the derivative of the circle's equation can be used to find the slope.

To find the slope of the tangent at a specific point, such as \((2 + \sqrt{3}, 3)\), implicit differentiation helps. Here, it is calculated as \(\frac{\sqrt{3}}{4}\).

This slope provides the necessary information to determine the direction vector of the tangent, which points along the path the circle rolls. The direction vector calculated can be any multiple that shows the proper orientation but is often simplified, like \((1, \sqrt{3})\), reflecting the slope's proportional relationship between horizontal and vertical changes.

Transformation of Coordinates

Coordinate transformation involves moving a shape within a plane without altering its size or shape. In exercises like the one discussed, it is essential when an object moves along a plane, such as when the circle rolls along its tangent. The direction vector guides this transformation, indicating the path the center of the circle should follow.

In this problem, after calculating the unit vector as \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\), the circle is rolled up by 2 units along its tangent direction. This transformation calculates to the new center position \((3, 4 + \sqrt{3})\).

Such calculations are essential for determining how the equation of a geometric figure changes after a transformation, ensuring correct adjustments to locate it within the coordinate plane. Understanding these principles of coordinate transformation allows one to navigate complex problems efficiently, adjusting centers and re-evaluating the position of figures correctly within the coordinate system.