The diameter of the circle, whose centre lies on the line x + y = 2 in the first quadrant and which

Q: Why is the radius $|\lambda - 3|$? Shouldn't it just be $3 - \lambda$?

That is a fantastic observation! When we calculate the distance from a point $(\lambda, k)$ to the vertical line $x = 3$, the distance is defined as $|\lambda - 3|$. Since we know $\lambda < 3$ (because the circle is in the first quadrant and touches $x=3$), the expression $\lambda - 3$ is indeed negative. To ensure our radius is a positive physical length, we must negate it, resulting in $3 - \lambda$. Always trust the modulus operator to guide you through the sign changes!

Q: Why does the center have to be $(\lambda, 2 - \lambda)$?

This is the power of parametric reduction! We have two variables, $h$ and $k$, but they are bound by the equation $h + k = 2$. By setting $h = \lambda$, we immediately know $k = 2 - \lambda$. This reduces a two-variable problem into a single-variable one, which is the hallmark of elegant problem solving. It simplifies the algebra significantly.

Q: Is it possible for the circle to be in another quadrant?

The problem specifically constrains the center to the first quadrant. This is a vital piece of information! It forces $\lambda > 0$ and $2 - \lambda > 0$. If we ignored this, we might find mathematically valid solutions that are physically impossible in the context of the problem. Always check your constraints before you start solving!

Question

The diameter of the circle, whose centre lies on the line $x + y = 2$ in the first quadrant and which touches both the lines $x = 3$ and $y = 2$, is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have a line, $x + y = 2$, cutting across the first quadrant like a diagonal bridge.

Somewhere on this bridge sits the center of a circle. This circle is constrained to touch two specific boundaries: the vertical wall $x = 3$ and the horizontal floor $y = 2$.

We define the center as $(h, k)$. Since it lies on the line $x + y = 2$, we use the power of parameterization. If we let the x-coordinate be $\lambda$, then the y-coordinate must be $2 - \lambda$.

Thus, our center is defined as $C(\lambda, 2 - \lambda)$. This substitution is the key that unlocks the entire problem.

### The Tangent Trap

Now, consider the lines $x = 3$ and $y = 2$. These are the tangents to the circle.

In geometry, the perpendicular distance from the center of a circle to any tangent line is exactly equal to the radius $r$. This is the fundamental property we must exploit.

For the vertical line $x = 3$, the distance is the absolute difference between the x-coordinate of the center and the line:
$$r = |\lambda - 3|$$

For the horizontal line $y = 2$, the distance is the absolute difference between the y-coordinate of the center and the line:
$$r = |(2 - \lambda) - 2|$$

This simplifies beautifully to $r = |-\lambda|$. Since we are in the first quadrant, we know $\lambda$ must be positive, so $|-\lambda| = \lambda$.

### The Algebraic Resolution

We now have two expressions for the same radius: $r = |\lambda - 3|$ and $r = \lambda$. Since a circle can only have one radius, we equate them:
$$|\lambda - 3| = \lambda$$

Because the circle is in the first quadrant and touches $x = 3$, $\lambda$ must be less than $3$. Therefore, the term $(\lambda - 3)$ is negative.

To remove the absolute value, we write:
$$-( \lambda - 3 ) = \lambda$$

This simplifies to $3 - \lambda = \lambda$. Solving this linear equation is straightforward:
$$2\lambda = 3 \Rightarrow \lambda = \frac{3}{2} = 1.5$$

### Final Calculation

We have found the center coordinate $\lambda = 1.5$. Consequently, the radius is $r = 1.5$.

The question asks for the diameter of the circle. The diameter $d$ is twice the radius:
$$d = 2 	imes 1.5 = 3$$

Through careful visualization and algebraic discipline, we have unraveled the mystery. The final answer is **3**.

Animated Solution for Mathematics - Circles: The diameter of the circle, whose centre lies on the line $x + y = 2$ in the first quadrant and which touches both the lines $x = 3$ and $y = 2$ , is

Enter Numerical Value:

Visualized Solution

Visualizing the Constraints

The Tangent Lines

Defining the Center $C (λ, 2 - λ)$

First Quadrant Constraint

Distance to the Line $x = 3$

Simplifying the First Radius

Distance to the Line $y = 2$

Simplifying the Second Radius

Equating the Radii

Solving for $λ$

Finding the Radius and Diameter

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Tangent Trap

The Algebraic Resolution

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Constraints

The Tangent Lines

Defining the Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C(λ,2−λ)

First Quadrant Constraint

Distance to the Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

Simplifying the First Radius

Distance to the Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=2

Simplifying the Second Radius

Equating the Radii

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding the Radius and Diameter

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Tangent Trap

The Algebraic Resolution

Final Calculation

Similar Questions

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