Problem 9 - A Complete Resource Book in Mathematics for JEE Main

Question

A square is inscribed in the circle \(x^{2}+y^{2}-2 x+4 y+3\) \(=0 .\) 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

The vertex of the square is (1+\sqrt{2}, -2).

Step 1: Identify the Circle's Center and Radius

The equation of the circle is given by \( x^2 + y^2 - 2x + 4y + 3 = 0 \). To find the center and radius, complete the square for both \( x \) and \( y \). - For \( x^2 - 2x \), add and subtract \( \left(\frac{-2}{2}\right)^2 = 1 \) to get \( (x-1)^2 - 1 \).- For \( y^2 + 4y \), add and subtract \( \left(\frac{4}{2}\right)^2 = 4 \) to get \( (y+2)^2 - 4 \).Rewriting the equation:\((x-1)^2 + (y+2)^2 = 1^2\)Hence, the center of the circle is \( (1, -2) \) and the radius is 1.

Step 2: Understand the Inscribed Square

A square is said to be inscribed in a circle if all its vertices touch the circle. Since the sides of the square are parallel to the coordinate axes, the diagonal of this square would be equal to the diameter of the circle, which is 2 (double the radius).

Step 3: Calculate the Side Length of the Square

The diagonal of the square is equal to the diameter of the circle. For a square with diagonal length \(d\), the side length \(s\) can be calculated using the formula \( s = \frac{d}{\sqrt{2}} \). Here, \( d = 2 \), so the side \( s = \frac{2}{\sqrt{2}} = \sqrt{2} \). Therefore, each side of the square is \( \sqrt{2} \).

Step 4: Determine the Vertex of the Square

Consider the center of the circle (1, -2) as the center of the square. Let's compute one of the vertices. One vertex of this square will be located at \((1 + \frac{\sqrt{2}}{2}, -2\)), which simplifies to \((1+\sqrt{2}, -2)\). The same computation can be done for each vertex, situated a distance \(\frac{\sqrt{2}}{2}\) from the center along both x and y-forming perpendicular directions, but here is only one such vertex to match the options.

Key Concepts

Circle Equations

Understanding the equation of a circle is fundamental in geometry and coordinate systems. The standard form of a circle's equation is \[ (x - h)^2 + (y - k)^2 = r^2 \]where

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

To find the center and radius of a given circle equation, it sometimes requires you to complete the square.
In the example exercise, the equation is \( x^2 + y^2 - 2x + 4y + 3 = 0 \).
We reshaped it into the standard form \((x-1)^2 + (y+2)^2 = 1^2\),revealing:

• The center is \((1, -2)\).
• The radius is 1.

Recognizing such transformations allows us to determine important properties about the circle, like the dimensions and locations of inscribed shapes.

Inscribed Shapes

In geometry, when a shape is inscribed in another, all its vertices touch the boundary of the larger shape. For instance, a square is inscribed in a circle if every corner of the square lies on the circle's circumference.
In an exercise scenario, the square is inscribed in a circle centered at \((1, -2)\) with radius 1.
The diagonal of the square equals the circle's diameter.
Given the circle's radius is 1, the diameter is twice that, so it's 2.
Knowing the square's diagonal, we can compute its side using the relation:\[ s = \frac{d}{\sqrt{2}} \]where \(d\) is the diagonal.
For this square, the side length is \(\sqrt{2}\),indicating each side is \(\sqrt{2}\) long.
This idea helps in problems involving inscribed squares or any polygons within circles.

Coordinate Geometry

Coordinate geometry merges algebra with geometry through the use of graphs and equations. It involves plotting shapes on a coordinate plane using mathematical equations to explore shapes' properties.
By mastering this, we solve various problems such as determining distances between points or understanding the layout of shapes.
One common problem uses coordinate geometry to find the square's position in relation to the circle.
With the square's center coinciding with the circle's center, calculations become easier.
For instance, to find a vertex of our inscribed square, add \(\frac{\sqrt{2}}{2}\) to both the horizontal and vertical span from the center:
From the example:

• The center of the circle and where the square centers is \((1, -2)\). Each side of the square is \(\sqrt{2}\) long.
• A vertex can be calculated by shifting from the circle center in both directions: \((1 + \sqrt{2}, -2)\) is a specific vertex given our calculations.

Coordinate geometry thus gives us powerful tools to understand and manipulate geometric configurations.