The centre of circle inscribed in square formed by the lines x² - 8x + 12 = 0 and y² - 14y + 45 = 0,

Q: Wait, why do these quadratic equations represent lines instead of curves?

That is a brilliant observation! Usually, $x^2$ implies a parabola. However, here the equations are independent. $x^2 - 8x + 12 = 0$ is a constraint on $x$ alone. It means that for *any* value of $y$, $x$ must be either $2$ or $6$. Thus, it defines the vertical lines $x=2$ and $x=6$.

Q: How do I know it is a square and not just a rectangle?

Great question! A rectangle becomes a square when its width equals its height. Here, the width is $|6 - 2| = 4$ and the height is $|9 - 5| = 4$. Since both dimensions are equal, the geometry is perfectly square.

Q: Do I need to find the radius of the circle to find the center?

Not at all! While the radius is indeed half the side length (which is $2$), the center of an inscribed circle is defined solely by the intersection of the lines of symmetry of the square. You only need the midpoints of the boundaries.

Q: Is there a shortcut if the lines were not parallel to the axes?

If the lines were slanted, we would need to find the intersection points of the four lines to identify the vertices of the square, then find the midpoint of the diagonals. But always look for symmetry first—it is the physicist's best friend!

Q: Why does the center of the square coincide with the center of the inscribed circle?

Because a square is a highly symmetric figure. The inscribed circle must touch all four sides at their midpoints. The point equidistant from all four sides is the geometric center, which is the intersection of the square's axes of symmetry.

Question

The centre of circle inscribed in square formed by the lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$, is

Options:

(A) $(4, 7)$

(B) $(7, 4)$

(C) $(9, 4)$

(D) $(4, 9)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty coordinate plane. You are given two cryptic clues: $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$.

At first glance, these look like standard quadratic equations. But in the realm of coordinate geometry, these are not just numbers; they are the blueprints for a geometric structure.

### Unveiling the Boundaries

Let us peel back the layers of the first equation: $x^2 - 8x + 12 = 0$. When we factorize this, we get $(x - 2)(x - 6) = 0$.

This tells us that $x$ must be either $2$ or $6$. In the Cartesian plane, $x=2$ is a vertical line passing through the point $(2, 0)$, and $x=6$ is another vertical line passing through $(6, 0)$. They stand like two parallel walls, exactly $4$ units apart.

Now, look at the second equation: $y^2 - 14y + 45 = 0$. Factoring this gives us $(y - 5)(y - 9) = 0$.

These represent two horizontal lines, $y=5$ and $y=9$. Just like the vertical lines, these are separated by a distance of $4$ units. When these four lines intersect, they trap a region of space. Because the distance between the vertical lines is equal to the distance between the horizontal lines, the shape formed is a **perfect square**.

### The Heart of the Square

We are tasked with finding the center of a circle inscribed within this square. Think of this circle as a perfectly fitting gear inside a square housing.

It must touch all four walls—$x=2, x=6, y=5,$ and $y=9$—at their exact midpoints. By the sheer elegance of symmetry, the center of this circle must be the center of the square itself.

To find this, we simply need the average of the boundaries. The $x$-coordinate of the center, $h$, is the midpoint of the vertical boundaries:

$$h = \frac{2 + 6}{2} = 4$$

Similarly, the $y$-coordinate, $k$, is the midpoint of the horizontal boundaries:

$$k = \frac{5 + 9}{2} = 7$$

### The Elegance of the Result

And there it is: **$(4, 7)$**. It feels almost too simple, but that is the beauty of geometry.

When you strip away the algebraic noise, you find that the problem was never about solving equations; it was about visualizing the symmetry of space. You have successfully navigated the boundaries, identified the square, and located its heart.

Keep this intuition sharp—in JEE Advanced, the most complex-looking problems often yield to the most elegant, symmetric solutions. You have mastered this one; now, carry that confidence forward.

Animated Solution for Mathematics - Straight Lines: The centre of circle inscribed in square formed by the lines $x^{2} - 8 x + 12 = 0$ and $y^{2} - 14 y + 45 = 0$ , is

Select Answer:

Visualized Solution

Analyze the Equation $x^{2} - 8 x + 12 = 0$

Identify Vertical Boundaries

Analyze the Equation $y^{2} - 14 y + 45 = 0$

Identify Horizontal Boundaries

The Formed Square

The Inscribed Circle

Center of the Circle (Logic)

Calculate the $x$ -coordinate

Calculate the $y$ -coordinate

Final Result: Center $(4, 7)$

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Unveiling the Boundaries

The Heart of the Square

The Elegance of the Result

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The centre of circle inscribed in square formed by the lines x2−8x+12=0 and y2−14y+45=0, is