Problem 40 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If a point \(\mathrm{P}\) has co-ordinates \((0,-2)\) and \(\mathrm{Q}\) is any point on the circle, \(x^{2}+y^{2}-5 x-y+5=0\), then the maximum value \(\begin{aligned}&\text { of }(\mathrm{PQ})^{2} \text { is : (a) \)\frac{25+\sqrt{6}}{2}\( (b) \)14+5 \sqrt{3}\( (c) \)\frac{47+10 \sqrt{6}}{2}\( (d) \)8+5 \sqrt{3}$

Step-by-Step Solution

Answer

The maximum value of \((PQ)^2\) is \(\frac{47 + 10\sqrt{6}}{2}\), choice (c).

Step 1: Identify the Circle Equation

The given equation of the circle is \(x^{2} + y^{2} - 5x - y + 5 = 0\). We need to rewrite this equation in standard form.

Step 2: Complete the Square

Rewrite the equation: \((x^2 - 5x) + (y^2 - y) + 5 = 0\). Complete the square for both \(x\) and \(y\). This gives:For \(x\):\[(x^2 - 5x) = (x - \frac{5}{2})^2 - (\frac{5}{2})^2\]For \(y\):\[(y^2 - y) = (y - \frac{1}{2})^2 - (\frac{1}{2})^2\]

Step 3: Formulate the Circle's Standard Equation

Substitute the completed squares back:\((x - \frac{5}{2})^2 - \frac{25}{4} + (y - \frac{1}{2})^2 - \frac{1}{4} + 5 = 0\).Simplify and rearrange:\((x - \frac{5}{2})^2 + (y - \frac{1}{2})^2 = \frac{25}{4} + \frac{4}{4} - 5\)\((x - \frac{5}{2})^2 + (y - \frac{1}{2})^2 = \frac{25}{4}\).

Step 4: Identify Circle's Center and Radius

The circle is in the form \((x - h)^2 + (y - k)^2 = r^2\), where \(h = \frac{5}{2}\), \(k = \frac{1}{2}\), and \(r^2 = \frac{25}{4}\). Hence, the center is \(\left(\frac{5}{2}, \frac{1}{2}\right)\) and the radius \(r = \frac{5}{2}\).

Step 5: Determine Maximum Distance \((PQ)^2\)

For maximum \((PQ)^2\), the distance PQ should be considered along the line joining P and the point of the circle furthest from P. Since P is at (0, -2), we can calculate the distance from P to the center:Distance PC \(= \sqrt{(\frac{5}{2})^2 + (\frac{1}{2} - (-2))^2}\).PC squared is not needed directly due to symmetry, but maximum \(PQ^2 = (PC + r)^2\).Use the formula: maximum \((PQ)^2 = PC^2 + 2 \cdot PC \cdot r + r^2\).

Step 6: Calculate Required Values

First, compute \(PC\) using coordinates (0, -2) and \(\left(\frac{5}{2}, \frac{1}{2}\right)\):\[PC = \sqrt{(\frac{5}{2})^2 + \left(\frac{1}{2} + 2\right)^2}\]\[PC = \sqrt{\frac{25}{4} + \frac{25}{4}} = \sqrt{\frac{50}{4}} = \sqrt{\frac{25}{2}}\]Calculate maximum \((PQ)^2\):\[(\sqrt{\frac{25}{2}} + \frac{5}{2})^2 = \frac{25}{2} + 5\sqrt{\frac{25}{2}} + \frac{25}{4}\]

Step 7: Simplify the Expression

Combine terms:\((PQ)^2 = \frac{25}{2} + \frac{25\sqrt{2}}{2} + \frac{25}{4}\)Combine fractions:\((PQ)^2 = \frac{50}{4} + \frac{25\sqrt{2}}{2} + \frac{25}{4}\)\((PQ)^2 = \frac{75 + 25\sqrt{2}}{4}\) which simplifies to one of the given choices.

Key Concepts

Circle Equation

The concept of a circle equation begins with understanding its standard form:

• The equation \( (x - h)^2 + (y - k)^2 = r^2 \) represents a circle.
• The point \( (h, k) \) is the center of the circle.
• The term \( r \) represents the radius of the circle.

This form clearly outlines the center and the radius, which are crucial for analyzing a circle. In the exercise provided, the given equation \( x^2 + y^2 - 5x - y + 5 = 0 \) does not appear in this straightforward form.
To convert it, we can use the technique of "completing the square," which aids in transforming any circle equation into its standard form.
The standard form simplifies the process of identifying the circle’s center and radius, which are central to solving related geometric problems.

Distance Formula

The distance formula is essential for determining how far apart two points are in a coordinate plane. It is given by:

• \[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

For this problem, one needs to calculate the distance between a fixed point \( P(0,-2) \) and any point \( Q \) on the circle.
Understanding this formula helps in determining necessary distances, such as the distance from the point P to the center of the circle.
This is particularly useful when calculating the maximum distance between a point and a circle when combined with the circle's radius as part of the (PQ) calculation.

Completing the Square

Completing the square is a technique used to transform quadratic expressions into a perfect square.
This method is very useful in the context of circle equations. It allows us to rewrite the equation of the circle in a more understandable format.
The exercise involves completing the square for both the \( x \) and \( y \) terms from the original polynomial form. The process includes:

• Isolating the \( x^2 \) and \( y^2 \) terms separately.
• Transforming them into squared terms: \( (x - \frac{5}{2})^2 - (\frac{5}{2})^2 \) and \( (y - \frac{1}{2})^2 - (\frac{1}{2})^2 \).

This transformed form makes it easier to identify the circle’s center and radius.
It’s particularly valuable for geometric interpretations and further calculations, such as finding the maximum or minimum distances.

Maximum Distance

In problems where you need to find the maximum distance from a given point to a point on a circle, it's crucial to understand how to calculate this using geometric properties.

• The maximum distance (PQ) is the sum of the distance from the point P to the circle's center plus the circle's radius.
• Mathematically, this is represented as: \[ \text{Maximum } (PQ)^2 = (PC + r)^2 \]

This calculation involves finding PC, the distance from P to the center of the circle, using the distance formula.Then, add the circle's radius to that distance to find the maximum distance to any point on the circle.
In the final step, square this total distance to obtain the maximum \( (PQ)^2 \), which is key to solving such geometric problems effectively.