JEE Main 2019 (11 January)

LEVELJEE Main

Animated Solution for Mathematics - Circles: A square is inscribed in the circle $x^{2} + y^{2} - 6 x + 8 y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is :

Q: Wait, why is the center of the circle the same as the center of the square? Is that always true?

That is a fantastic observation! Yes, for a square inscribed in a circle, the center of the circle is the point of symmetry for the square. Since the square's sides are parallel to the axes, the diagonals of the square are diameters of the circle, intersecting exactly at the circle's center. It is a beautiful property of symmetry.

Q: Why can't we just find the distance from the center to the origin and subtract the radius?

That would only work if the origin, the center of the circle, and the nearest vertex were collinear. However, the vertices are at specific coordinates $(x \pm \frac{a}{2}, y \pm \frac{a}{2})$. The distance from the origin to the vertex is not simply $OC - R$. We must calculate the distance to each vertex individually to be certain.

Q: Is there a shortcut to avoid calculating all four distances?

While you could sketch the circle and the square to visually estimate which vertex is closest to the origin, in a JEE Advanced scenario, calculating all four is safer and faster than risking a geometric misinterpretation. It takes only a few seconds to compute the squares of the coordinates!

Select Answer:

Visualized Solution

Analyze the Circle Equation

Given circle: $x^{2} + y^{2} - 6 x + 8 y - 103 = 0$

Standard Form of Circle

Convert to standard form: $(x - h)^{2} + (y - k)^{2} = R^{2}$

Completing the Square

Group terms: $(x^{2} - 6 x) + (y^{2} + 8 y) = 103$

Center and Radius

$(x^{2} - 6 x + 9) + (y^{2} + 8 y + 16) = 103 + 9 + 16$
$(x - 3)^{2} + (y + 4)^{2} = 128$
Center $C (3, - 4)$ , Radius $R = 128 = 82$

Geometry of Inscribed Square

Square is inscribed in the circle.
Diagonal of square = Diameter of circle = $2 R$

Side Length of Square

Diameter $= 2 \times 82 = 162$
Let side length be $a$ . Diagonal $= a 2$
$a 2 = 162 ⟹ a = 16$

Locating the Vertices

Sides are parallel to coordinate axes.
Distance from center to any side is $\frac{a}{2} = 8$
Vertices: $(x_{c} \pm 8, y_{c} \pm 8)$

Coordinates of Vertices

Center $(x_{c}, y_{c}) = (3, - 4)$
Vertices are at $(3 \pm 8, - 4 \pm 8)$

Calculating the Four Vertices

$V_{1} = (11, 4)$
$V_{2} = (11, - 12)$
$V_{3} = (- 5, 4)$
$V_{4} = (- 5, - 12)$

Distance from Origin

Distance from $(0, 0)$ to $(x, y)$ is $d = x^{2} + y^{2}$
We need the minimum distance.

Evaluating Distances

$d_{1} = 1 1^{2} + 4^{2} = 137$
$d_{2} = 1 1^{2} + (- 12)^{2} = 265$
$d_{3} = (- 5)^{2} + 4^{2} = 41$
$d_{4} = (- 5)^{2} + (- 12)^{2} = 13$

The Nearest Vertex

Comparing: $41 < 137 < 13 < 265$
Minimum distance is $41$
Nearest vertex is $(- 5, 4)$

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Circle

We begin with the equation of a circle:

x^{2} + y^{2} - 6 x + 8 y - 103 = 0

. To reveal its properties, we transform this into the standard form

(x - h)^{2} + (y - k)^{2} = R^{2}

By grouping the

x

and

y

terms, we write:

(x^{2} - 6 x) + (y^{2} + 8 y) = 103

Completing the square, we add

9

(from

(- 3)^{2}

) and

16

(from

4^{2}

) to both sides:

(x - 3)^{2} + (y + 4)^{2} = 103 + 9 + 16

This simplifies to the standard form:

(x - 3)^{2} + (y + 4)^{2} = 128

The center of the circle is

C (3, - 4)

and the radius is

R = 128 = 82

The Inscribed Square

For a square inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. The diameter is

D = 2 R = 162

If the side length of the square is

a

, then the diagonal is

a 2

. Equating these:

a 2 = 162 \Rightarrow a = 16

Since the sides are parallel to the coordinate axes, the distance from the center

(3, - 4)

to any side is exactly half the side length, which is

\frac{16}{2} = 8

Calculating the Vertices

To find the vertices, we move

8

units horizontally and

8

units vertically from the center

C (3, - 4)

. This yields the four vertices:

V = (3 \pm 8, - 4 \pm 8)

The specific coordinates are:

V_{1} = (11, 4)

V_{2} = (11, - 12)

V_{3} = (- 5, 4)

V_{4} = (- 5, - 12)

Determining the Nearest Vertex

We calculate the distance

d

from the origin

(0, 0)

to each vertex using

d = x^{2} + y^{2}

For

V_{1}

d = 1 1^{2} + 4^{2} = 121 + 16 = 137

For

V_{2}

d = 1 1^{2} + (- 12)^{2} = 121 + 144 = 265

For

V_{3}

d = (- 5)^{2} + 4^{2} = 25 + 16 = 41

For

V_{4}

d = (- 5)^{2} + (- 12)^{2} = 25 + 144 = 169 = 13

Comparing these values,

41

is the smallest distance. The nearest vertex is at a distance of $41$ .

Common Traps & FAQs

Wait, why is the center of the circle the same as the center of the square? Is that always true?

That is a fantastic observation! Yes, for a square inscribed in a circle, the center of the circle is the point of symmetry for the square. Since the square's sides are parallel to the axes, the diagonals of the square are diameters of the circle, intersecting exactly at the circle's center. It is a beautiful property of symmetry.

Why can't we just find the distance from the center to the origin and subtract the radius?

That would only work if the origin, the center of the circle, and the nearest vertex were collinear. However, the vertices are at specific coordinates

(x \pm \frac{a}{2}, y \pm \frac{a}{2})

. The distance from the origin to the vertex is not simply

O C - R

. We must calculate the distance to each vertex individually to be certain.

Is there a shortcut to avoid calculating all four distances?

While you could sketch the circle and the square to visually estimate which vertex is closest to the origin, in a JEE Advanced scenario, calculating all four is safer and faster than risking a geometric misinterpretation. It takes only a few seconds to compute the squares of the coordinates!

Animated Solution for Mathematics - Circles: A square is inscribed in the circle $x^{2} + y^{2} - 6 x + 8 y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is :

Select Answer:

Visualized Solution

Analyze the Circle Equation

Standard Form of Circle

Completing the Square

Center and Radius

Geometry of Inscribed Square

Side Length of Square

Locating the Vertices

Coordinates of Vertices

Calculating the Four Vertices

Distance from Origin

Evaluating Distances

The Nearest Vertex

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Circle

The Inscribed Square

Calculating the Vertices

Determining the Nearest Vertex

Similar Questions

A square is inscribed in the circle $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$ . One side of this square is parallel to $y = x + 3$ . If $(x_{i}, y_{i})$ are the vertices of the square, then $\sum (x_{i}^{2} + y_{i}^{2})$ is equal to:

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

If the area of an equilateral triangle inscribed in the circle, $x^{2} + y^{2} + 10 x + 12 y + c = 0$ is $273$ sq. units then c is equal to :

If the circles $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ and $x^{2} + y^{2} + 2 (3 - 3) x + 2 (4 - 6) y = k + 63 + 86, k > 0$ , touch internally at the point $P (α, β)$ , then $(α + 3)^{2} + (β + 6)^{2}$ is equal to ________

Let the circle $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$ and $C_{2}$ be a circle having centre at $(- 1, 0)$ and radius 2. If the line of the common chord of $C_{1}$ and $C_{2}$ intersects the $y$ -axis at the point $P$ , then the square of the distance of $P$ from the centre of $C_{1}$ is :

The minimum distance between any two points $P_{1}$ and $P_{2}$ while considering point $P_{1}$ on one circle and point $P_{2}$ on the other circle for the given circles' equations $x^{2} + y^{2} - 10 x - 10 y + 41 = 0$ and $x^{2} + y^{2} - 24 x - 10 y + 160 = 0$ is ____.

The equation of a circle is $Re (z^{2}) + 2 (Im (z))^{2} + 2 Re (z) = 0$ , where $z = x + i y$ . A line which passes through the center of the given circle and the vertex of the parabola, $x^{2} - 6 x - y + 13 = 0$ , has $y$ -intercept equal to ___

A rectangle $R$ with end points of the one of its sides as $(1, 2)$ and $(3, 6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2 x - y + 4 = 0$ , then the area of $R$ is _______.

Let $P$ and $Q$ be two distinct points on a circle which has center at $C (2, 3)$ and which passes through origin $O$ . If $O C$ is perpendicular to both the line segments $C P$ and $C Q$ , then the set ${P, Q}$ is equal to

Select Answer:

Visualized Solution

Analyze the Circle Equation

Standard Form of Circle

Completing the Square

Center and Radius

Geometry of Inscribed Square

Side Length of Square

Locating the Vertices

Coordinates of Vertices

Calculating the Four Vertices

Distance from Origin

Evaluating Distances

The Nearest Vertex

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Circle

The Inscribed Square

Calculating the Vertices

Determining the Nearest Vertex

Similar Questions

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the area of an equilateral triangle inscribed in the circle, x2+y2+10x+12y+c=0 is 273​ sq. units then c is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the circles x2+y2+6x+8y+16=0 and x2+y2+2(3−3​)x+2(4−6​)y=k+63​+86​,k>0, touch internally at the point P(α,β), then (α+3​)2+(β+6​)2 is equal to ________

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

If the area of an equilateral triangle inscribed in the circle, $x^{2} + y^{2} + 10 x + 12 y + c = 0$ is $273$ sq. units then c is equal to :

If the circles $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ and $x^{2} + y^{2} + 2 (3 - 3) x + 2 (4 - 6) y = k + 63 + 86, k > 0$ , touch internally at the point $P (α, β)$ , then $(α + 3)^{2} + (β + 6)^{2}$ is equal to ________