Problem 129 - A Complete Resource Book in Mathematics for JEE Main

Question

The length of the diameter of the circle which touches the \(x\)-axis at the point \((1,0)\) and passes through the point \((2,3)\) is [2012] (A) \(\frac{10}{3}\) (B) \(\frac{3}{5}\) (C) \(\frac{6}{5}\) (D) \(\frac{5}{3}\)

Step-by-Step Solution

Answer

The diameter of the circle is \(\frac{10}{3}\).

Step 1: Understand the Problem

We need to find the diameter of a circle that touches the \(x\)-axis at \((1,0)\) and passes through the point \((2,3)\). Touching the \(x\)-axis means the \(x\)-axis is tangent to the circle, and \((1,0)\) is the point of tangency.

Step 2: Geometry Insight

For a circle touching the \(x\)-axis at \((1,0)\), the center must be directly above or below the \(x\)-axis in line with this point. Hence, the center of the circle can be \((1, r)\) or \((1, -r)\), where \(r\) is the radius of the circle.

Step 3: Applying the Point to the Circle Equation

Since the circle passes through \((2,3)\), use the distance formula to calculate \(r\). The distance from the center \((1, r)\) to \((2, 3)\) must be equal to \(r\). Thus, \((2-1)^2 + (3-r)^2 = r^2\).

Step 4: Solve the Radius Equation

Simplify the equation: \[1 + (3-r)^2 = r^2\] Expand and simplify further: \[1 + 9 - 6r + r^2 = r^2\]Thus, we have \[10 - 6r = 0\]. So, \(r = \frac{10}{6} = \frac{5}{3}\).

Step 5: Determine the Diameter

The diameter of the circle is twice the radius, so it is \[2 \times \frac{5}{3} = \frac{10}{3}\].

Key Concepts

Circle Equation

The equation of a circle is a fundamental concept in geometry. It helps us understand the properties and locus of points on a plane that form a circle.
The standard form of a circle's equation is given by \[(x - h)^2 + (y - k)^2 = r^2\]where:

• \( (h, k)\) is the center of the circle.
• \( r \) is the radius of the circle.

For a specific problem, once we know the center and radius, we can completely determine the circle's equation. This allows us to analyze relationships between different points and how they lie on or off the circle.

Radius and Diameter

The radius and diameter are essential components in understanding the size of a circle.
The radius \( r \) of a circle is the distance from the center of the circle to any point on the circle. In this problem, the radius was found using the distance formula by measuring from the circle's center to a given point on its circumference.
Remember, the formula used was the distance equation: \[(x_2 - x_1)^2 + (y_2 - y_1)^2 = r^2\]
The diameter is twice the length of the radius, so if you know the radius, just multiply it by 2 to find the diameter!
In this exercise, knowing how to manipulate these simple formulas helped us to find the diameter correctly.

Tangent to Axis

A tangent to a circle is a line that touches the circle at exactly one point. In the given problem, the circle touches the \(x\)-axis, making the \(x\)-axis a tangent to the circle.
When a circle touches an axis, it means the center of the circle is vertically aligned above or below the tangency point.
For example, if a circle is tangential to the \(x\)-axis at point \((1, 0)\), then the center of the circle must have the coordinates \((1, r)\) or \((1, -r)\) where \(r\) is the radius.
Recognizing tangency conditions provides crucial insight because it simplifies locating potential circle centers with respect to the axis, ultimately aiding in solving circle geometry problems like this one.