Problem 66 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If a square is inscribed in the circle \(x^{2}+y^{2}-2 x+4 y+\) \(3=0\) and its sides are parallel to the coordinate axes, then one vertex of the square is (A) \((1+\sqrt{2},-2)\) (B) \((1-\sqrt{2},-2)\) (C) \((1,-2+\sqrt{2})\) (D) none of these

Step-by-Step Solution

Answer

None of these

Step 1: Rewrite Circle Equation to Standard Form

Given the circle equation: \(x^{2} + y^{2} - 2x + 4y + 3 = 0\). To convert to standard form, complete the square for both \(x\) and \(y\). For \(x\), rearrange terms: \(x^2 - 2x\). Completing the square: \(x^2 - 2x = (x-1)^2 - 1\). For \(y\), rearrange terms: \(y^2 + 4y\). Completing the square: \(y^2 + 4y = (y+2)^2 - 4\). Substitute back into the original equation: \((x-1)^2 - 1 + (y+2)^2 - 4 + 3 = 0\). Simplify: \((x-1)^2 + (y+2)^2 = 2\). So the circle center is \((1, -2)\) with radius \(\sqrt{2}\).

Step 2: Determine the Square's Vertices Coordinates

The square is inscribed in the circle with its sides parallel to the coordinate axes. Hence, its center is the circle's center \((1, -2)\). For a square inscribed in a circle, the diagonal of the square equals the circle's diameter, which is \(2r\). Therefore, diagonal \(= 2\sqrt{2}\). Let the side of the square be \(a\), then the diagonal: \(a\sqrt{2} = 2\sqrt{2}\), giving \(a = 2\). Each vertex of the square lies \(\pm a/2\) from the center: \( (1+1, -2+1), (1-1, -2+1), (1+1, -2-1), (1-1, -2-1) \).

Step 3: Verify the Calculated Vertices Coordinates

The vertex coordinates from Step 2 are \((2, -1), (0, -1), (2, -3), (0, -3)\). Compare these with the given options. None of these match these calculated coordinates \((2, -1), (0, -1), (2, -3), (0, -3)\). To verify, none match \((1+\sqrt{2}, -2), (1-\sqrt{2}, -2), (1, -2+\sqrt{2})\). Hence, the correct answer is "none of these".

Key Concepts

Circle Equation

The circle equation in a geometric problem often starts in a general quadratic form. To understand the properties of the circle, such as its center and radius, we typically need to transform this equation into the standard form. The general form of a circle's equation looks like this:

• \[ x^2 + y^2 + Dx + Ey + F = 0 \]

To convert it into the standard form, which is

• \[ (x-h)^2 + (y-k)^2 = r^2 \]

we must complete the square for both the \(x\) and \(y\) terms. Let's break this down:- For the \(x\) terms in the equation \(x^2 - 2x\), completing the square means rewriting it as \((x-1)^2 - 1\).- Similarly, for \(y^2 + 4y\), we rewrite it as \((y+2)^2 - 4\). This process reveals important characteristics:

• The circle's center is at \((h, k)\), here \((1, -2)\).
• The circle's radius, \(r\), can be calculated from the equation \( (x-1)^2 + (y+2)^2 = r^2 \), giving us a radius \(r = \sqrt{2}\).

Understanding circle equations helps in solving problems involving shape geometry and positions.

Square Inscription

When a square is inscribed within a circle, it means that the vertices of the square touch the circle, and the circle passes through each corner of the square. Let’s delve into the geometry:- In such a configuration, the diagonal of the square is equal to the diameter of the circle, which allows us to calculate the side length of the square.- For this given problem, the circle has a radius of \(\sqrt{2}\), which means the diameter is \(2\sqrt{2}\).- The square's diagonal being \(2\sqrt{2}\) leads us to set up the relationship for the square's side length \(a\): \[ a\sqrt{2} = 2\sqrt{2} \] Simplifying gives us, \(a = 2\).Position the square such that its sides are parallel to the coordinate axes makes calculations straightforward, as it maintains symmetry around the circle center, here at \((1, -2)\). Each vertex is positioned at distances ±1 vertically or ±1 horizontally from the circle center, forming a perfectly positioned inscribed square.

Standard Form Conversion

The standard form conversion is crucial in understanding and visualizing the geometrical figures represented by an equation. When dealing with equations of circles, this process of converting the given equation to standard form allows us to quickly identify the center and the radius, which are vital for any geometry problem.- Begin by completing the square for both the \(x\) and \(y\) terms separately. We illustrated this in the earlier section, moving from \[ x^2 + y^2 - 2x + 4y + 3 = 0 \] to the standard circle form: \[ (x-1)^2 + (y+2)^2 = 2 \]- Rearranging and simplifying helps you extract the circle's center. In this instance, it's \((1, -2)\).- It also reveals the circle's radius square, \(2\), indicating a radius of \(\sqrt{2}\).Such conversions are beneficial in geometry to plan and execute further operations, whether understanding relations with other shapes like inscribed squares, or merely discerning the circle's size and placement in the coordinate plane. Understanding how each part of the original equation shifts into place during the transformation is essential for mastering coordinate geometry.