A square is inscribed in the circle x² + y² - 2x + 4y + 3 = 0. Its sides are parallel to the coordin

Q: Wait, why is the center of the square the same as the center of the circle?

Great observation! Because the square is inscribed in the circle and its sides are parallel to the coordinate axes, the square inherits the symmetry of the circle. Any square inscribed in a circle must have its center coincide with the circle's center. If it were shifted, the vertices would not all touch the circle's circumference simultaneously.

Q: How do I know the vertices are $(h \pm a, k \pm a)$?

Think about the geometry! If the sides are parallel to the axes, the square is essentially a box aligned with the grid. If the center is $(h, k)$ and the side length is $2a$, then moving from the center to the right edge means adding $a$ to the x-coordinate, and moving to the top edge means adding $a$ to the y-coordinate. Thus, the top-right vertex is $(h+a, k+a)$. By symmetry, the other vertices are just reflections across the center.

Q: I keep getting 'none of these' as the answer. Is it possible that I calculated the radius wrong?

It is always good to double-check! Let's re-verify: the circle equation is $x^2 + y^2 - 2x + 4y + 3 = 0$. Completing the square gives $(x-1)^2 + (y+2)^2 = 2$. So, $r^2 = 2$, which means $r = \sqrt{2}$. Your calculation is likely correct. In JEE Advanced, 'none of these' is a perfectly valid and sometimes necessary answer. Trust your derivation!

Question

A square is inscribed in the circle $x^2 + y^2 - 2x + 4y + 3 = 0$. Its sides are parallel to the coordinate axes. One vertex of the square is

Options:

(A) $(1 + \sqrt{2}, -2)$

(B) $(1 - \sqrt{2}, -2)$

(C) $(1, -2 + \sqrt{2})$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Circle

Every journey begins with a map. In our case, the map is the equation of the circle: 
$$x^2 + y^2 - 2x + 4y + 3 = 0$$

Before we can analyze the square, we must determine the circle's center and radius. We compare the given equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

By comparing coefficients, we find $2g = -2$ and $2f = 4$, which yields the center $(h, k) = (-g, -f) = (1, -2)$.

To find the radius, we use the formula $r = \sqrt{g^2 + f^2 - c}$. Substituting our values:
$$r = \sqrt{(-1)^2 + (2)^2 - 3} = \sqrt{1 + 4 - 3} = \sqrt{2}$$

We now know exactly where our circle is positioned and its extent.

### The Symmetry of the Square

The square is inscribed, meaning all four of its vertices lie on the circle. Crucially, the problem states its sides are parallel to the coordinate axes.

Because the square is perfectly aligned with the axes, its center must be identical to the center of the circle, $(1, -2)$. If the square were tilted, this property would not hold, but here, the symmetry is absolute.

### The Pythagorean Bridge

Let the distance from the center of the square to any of its sides be $a$. Consequently, the side length of the square is $2a$.

If we consider the top-right vertex, its coordinates relative to the center $(1, -2)$ are $(1+a, -2+a)$. We know that the distance from the center to any vertex is equal to the radius of the circle, $r = \sqrt{2}$.

We form a right-angled triangle where the base is $a$, the height is $a$, and the hypotenuse is the radius $r$. By the Pythagorean theorem:
$$a^2 + a^2 = r^2$$

Substituting our radius value:
$$2a^2 = (\sqrt{2})^2$$
$$2a^2 = 2$$
$$a^2 = 1 \Rightarrow a = 1$$

### The Final Revelation

With $a=1$, we can pinpoint the vertices. Starting from the center $(1, -2)$, we move $1$ unit in each direction to find the corners:

$$(1+1, -2+1) = (2, -1)$$
$$(1+1, -2-1) = (2, -3)$$
$$(1-1, -2+1) = (0, -1)$$
$$(1-1, -2-1) = (0, -3)$$

The vertices of the square are $(2, -1), (2, -3), (0, -1),$ and $(0, -3)$.

If these coordinates do not match the provided options, do not panic. In the context of JEE, **"none of these"** is a valid outcome that confirms you have navigated the geometry correctly. You have successfully mastered the problem.

Animated Solution for Mathematics - Circles: 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. One vertex of the square is

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Visualized Solution

The Circle Equation $x^{2} + y^{2} - 2 x + 4 y + 3 = 0$

General Equation of a Circle

Extracting $g$ , $f$ , and $c$

Calculating the Center $(h, k)$

Formula for the Radius $r$

Calculating the Radius $r = 2$

Inscribing the Square

Connecting Center to Vertex

Geometry of the Square

Applying Pythagoras Theorem

Solving for $k$

Coordinates of Vertices

Finding the Exact Vertices

Checking the Options

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Circle

The Symmetry of the Square

The Pythagorean Bridge

The Final Revelation

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A square is inscribed in the circle x2+y2−2x+4y+3=0. Its sides are parallel to the coordinate axes. One vertex of the square is