Problem 26 - A Complete Resource Book in Mathematics for JEE Main

Question

The coordinates of the point on the circle \(x^{2}+y^{2}-2 x\) \(-4 y-11=0\) farthest from the origin are (A) \(\left(2+\frac{8}{\sqrt{5}}, 1+\frac{4}{\sqrt{5}}\right)\) (B) \(\left(1+\frac{4}{\sqrt{5}}, 2+\frac{8}{\sqrt{5}}\right)\) (C) \(\left(1+\frac{8}{\sqrt{5}}, 2+\frac{4}{\sqrt{5}}\right)\) (D) none of these

Step-by-Step Solution

Answer

The farthest point is \((1 - \frac{4}{\sqrt{5}}, 2 - \frac{8}{\sqrt{5}})\). Answer: (D) none of these.

Step 1: Convert Circle to Standard Form

The given equation of the circle is \(x^2 + y^2 - 2x - 4y - 11 = 0\). To convert it to standard form, complete the square for both \(x\) and \(y\) terms. First, rearrange it as \((x^2 - 2x) + (y^2 - 4y) = 11\).

Step 2: Complete the Square for x

Take \(x^2 - 2x\) and complete the square by adding and subtracting \(1\) inside the equation. Thus, \(x^2 - 2x + 1 - 1 = (x - 1)^2 - 1\).

Step 3: Complete the Square for y

Take \(y^2 - 4y\) and complete the square by adding and subtracting \(4\). Thus, \(y^2 - 4y + 4 - 4 = (y - 2)^2 - 4\).

Step 4: Rewrite the Circle in Standard Form

Substitute the completed squares back into the left-hand side: \((x - 1)^2 - 1 + (y - 2)^2 - 4 = 11\). This simplifies to \((x - 1)^2 + (y - 2)^2 = 16\).

Step 5: Identify the Center and Radius

The standard form equation \((x - 1)^2 + (y - 2)^2 = 16\) shows that the center of the circle is \((1, 2)\) and the radius is \(\sqrt{16} = 4\).

Step 6: Find the Farthest Point from Origin

The point on the circle that is farthest from the origin will lie along the line connecting the origin and the center of the circle, extended out by the radius. Calculate the direction vector from the center \((1, 2)\) to the origin \((0, 0)\). This vector is \((-1, -2)\).

Step 7: Calculate the Farthest Point

Normalize the direction vector \((-1, -2)\) to have length one: divide by its magnitude \(\sqrt{(-1)^2 + (-2)^2} = \sqrt{5}\) to get \((-\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}})\). Scale this vector by the circle's radius, which is 4: \((-\frac{4}{\sqrt{5}}, -\frac{8}{\sqrt{5}})\). Add this vector to the center \((1, 2)\) to obtain the farthest point: \((1 - \frac{4}{\sqrt{5}}, 2 - \frac{8}{\sqrt{5}})\).

Step 8: Compare with Options

The calculated farthest point is \((1 - \frac{4}{\sqrt{5}}, 2 - \frac{8}{\sqrt{5}})\). None of the options \((A), (B), (C)\) match this point. Thus, the correct answer is none of these.

Key Concepts

Completing the Square

Completing the square is a vital technique in algebra that allows us to transform a quadratic equation into a perfect square trinomial. This technique simplifies equations and makes them easier to work with, particularly when dealing with quadratic terms.
In the context of circle geometry, we often use it to convert the general form of a circle's equation into its standard form. When we have quadratic terms like \(x^2 - 2x\) or \(y^2 - 4y\), completing the square involves:

• Identifying the linear term (e.g., -2x or -4y).
• Halving the coefficient of the linear term and squaring the result.
• Adding and subtracting this square inside the equation to maintain equality.

For instance, with \(x^2 - 2x\), taking half of -2 gives -1, and squaring it results in 1. Hence, we adjust \(x^2 - 2x\) to \((x-1)^2 - 1\). Likewise, \(y^2 - 4y\) becomes \((y-2)^2 - 4\) after completing the square. These transformations allow us to express the circle's equation in a more manageable and recognizable standard form.

Standard Form of a Circle

The standard form of a circle's equation is incredibly useful for identifying key attributes such as the center and radius of the circle. This form appears as \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) represents the center of the circle and \(r\) is the radius.
To convert an equation to this standard form, completing the square is essential. Once completed, the example equation \((x - 1)^2 + (y - 2)^2 = 16\) reveals that our circle has its center at \((1, 2)\) and a radius of 4, since \(16\) is \(4^2\).
Understanding this form is beneficial because it simplifies the process of interpreting an equation's geometric representation. Additionally, it sets the foundation for solving more complex problems that involve a circle's relationship with points and lines in the plane.

Farthest Point from Origin

In geometry, finding the farthest point from the origin on a circle involves understanding the orientation and dimensions of the circle relative to the origin. The farthest point will lie opposite the direction towards the origin, moving outward from the center by a distance equivalent to the radius.
The core steps involve:

• Identifying the direction vector from the circle's center to the origin. In our example, this is \((-1, -2)\) from \((1, 2)\) to \((0, 0)\).
• Normalizing this vector to ensure it has a magnitude of 1, turning it into unit form: \((\frac{-1}{\sqrt{5}}, \frac{-2}{\sqrt{5}})\).
• Scaling this unit vector by the circle's radius, here 4, leads to \((\frac{-4}{\sqrt{5}}, \frac{-8}{\sqrt{5}})\).
• Adding this vector to the center to determine the farthest point, resulting in \((1 - \frac{4}{\sqrt{5}}, 2 - \frac{8}{\sqrt{5}})\).

This systematic method provides us with the point that is the greatest distance from the origin, an important concept when analyzing spatial relationships within circle geometry.