The circle x² + y² - 4x - 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the

Q: Wait, why is the circumcenter the midpoint of the hypotenuse? Can't it be anywhere else?

That is a fantastic question! In any right-angled triangle, the hypotenuse acts as the diameter of the circumcircle. By definition, the center of a circle is the midpoint of its diameter. Therefore, for a right-angled triangle, the circumcenter must lie exactly at the midpoint of the hypotenuse. It is a beautiful, rigid geometric constraint that simplifies our life immensely.

Q: Why did we multiply by the conjugate? That feels like a random trick.

It is not random at all! We needed to transform an expression involving a square root into one involving an $xy$ product to match the form of the locus given in the problem. The conjugate $(x+y+\sqrt{x^2+y^2})$ is the perfect tool to rationalize the expression. When you multiply $(x+y-\sqrt{x^2+y^2})$ by $(x+y+\sqrt{x^2+y^2})$, you get $(x+y)^2 - (x^2+y^2)$, which expands to $x^2+y^2+2xy - x^2 - y^2$. The squares cancel out, leaving only the $2xy$ term. It is pure algebraic elegance!

Q: Why use the inradius formula $r = \frac{a+b-c}{2}$ instead of the general $Area = rs$ formula?

You could use the general formula, but it would be much slower. The formula $r = \frac{a+b-c}{2}$ is a specialized, high-efficiency tool derived specifically for right-angled triangles. In JEE Advanced, recognizing these specialized shortcuts is what separates the good from the great. It allows us to bypass calculating the semi-perimeter and area, getting us straight to the relationship between the legs and the hypotenuse.

Question

The circle $x^2 + y^2 - 4x - 4y + 4 = 0$ is inscribed in a triangle which has two of its sides along the co-ordinate axes. The locus of the circumcentre of the triangle is $x + y - xy + k(x^2 + y^2)^{1/2} = 0$. Find $k$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the coordinate plane. Today, we are not just solving a problem; we are embarking on a journey through the architecture of geometry.

We are given a circle, $x^2 + y^2 - 4x - 4y + 4 = 0$, nestled comfortably within a right-angled triangle whose sides are anchored to the coordinate axes. Our mission is to find the locus of the circumcenter of this triangle.

### Decoding the Circle

Before we touch the triangle, we must understand the circle. The equation $x^2 + y^2 - 4x - 4y + 4 = 0$ can be simplified by completing the square.

Grouping the $x$ terms and the $y$ terms, we get $(x^2 - 4x) + (y^2 - 4y) + 4 = 0$. Adding and subtracting the necessary constants, we transform this into:

$$(x-2)^2 - 4 + (y-2)^2 - 4 + 4 = 0$$

Simplifying this, we arrive at $(x-2)^2 + (y-2)^2 = 4$. This is the standard form of a circle equation, $(x-h)^2 + (y-k)^2 = r^2$.

Instantly, the fog clears: the center of our circle is at $(2, 2)$ and its radius $r$ is $2$. This circle is our anchor.

### The Geometry of the Triangle

Now, imagine the triangle. It is a right-angled triangle with its legs lying on the $x$ and $y$ axes. Let the vertices be the origin $O(0,0)$, point $A(a,0)$ on the $x$-axis, and point $B(0,b)$ on the $y$-axis.

The hypotenuse is the segment $AB$. Here is the crucial realization: the circumcenter $M(x,y)$ of a right-angled triangle is always the midpoint of the hypotenuse.

If $M$ is the midpoint of $AB$, then by the midpoint formula, $x = \frac{a}{2}$ and $y = \frac{b}{2}$. This gives us the vital link: $a = 2x$ and $b = 2y$.

### The Inradius Bridge

We need a relationship between the triangle's sides ($a$ and $b$) and the circle's radius ($r=2$). For a right-angled triangle, there is a powerful, specialized formula for the inradius:

$$r = \frac{a + b - \sqrt{a^2 + b^2}}{2}$$

Substituting our known values, we get:

$$2 = \frac{a + b - \sqrt{a^2 + b^2}}{2}$$

Now, we substitute our expressions for $a$ and $b$ in terms of $x$ and $y$:

$$2 = \frac{2x + 2y - \sqrt{(2x)^2 + (2y)^2}}{2}$$

Simplifying the square root, we get $\sqrt{4(x^2 + y^2)} = 2\sqrt{x^2 + y^2}$. The equation becomes:

$$2 = \frac{2x + 2y - 2\sqrt{x^2 + y^2}}{2}$$

Dividing by $2$, we obtain the fundamental relation:

$$2 = x + y - \sqrt{x^2 + y^2}$$

### The Algebraic Dance

We are close, but we must express this in a standard locus form. We rearrange the equation to isolate the square root:

$$\sqrt{x^2 + y^2} = x + y - 2$$

To resolve this, we multiply both sides of $2 = x + y - \sqrt{x^2 + y^2}$ by the conjugate $(x + y + \sqrt{x^2 + y^2})$. The left side becomes $2(x + y + \sqrt{x^2 + y^2})$.

The right side becomes the difference of squares:

$$(x + y)^2 - (\sqrt{x^2 + y^2})^2 = (x^2 + y^2 + 2xy) - (x^2 + y^2) = 2xy$$

Equating the two sides, we have:

$$2(x + y + \sqrt{x^2 + y^2}) = 2xy$$

Dividing by $2$, we get $x + y + \sqrt{x^2 + y^2} = xy$. Rearranging this, we arrive at the final locus:

**$$x + y - xy + \sqrt{x^2 + y^2} = 0$$**

Animated Solution for Mathematics - Circles: The circle $x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ is inscribed in a triangle which has two of its sides along the co-ordinate axes. The locus of the circumcentre of the triangle is $x + y - x y + k (x^{2} + y^{2})^{1/2} = 0$ . Find $k$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Analyze the Circle Equation

Identify Center and Radius

Define Triangle Vertices

Locate the Circumcenter

Relate Circumcenter to Vertices

The Inradius Property

Substitute Known Values

Substitute Locus Coordinates

Simplify the Equation

The Algebraic Trick

Execute the Multiplication

Simplify the Expression

Form the Final Locus

Compare and Conclude

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Decoding the Circle

The Geometry of the Triangle

The Inradius Bridge

The Algebraic Dance

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