Problem 131 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(C\) be the circle with centre at \((1,1)\) and with radius 1\. If \(T\) is the circle centered at \((0, y)\), passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal to (A) \(\frac{\sqrt{3}}{\sqrt{2}}\) (B) \(\frac{\sqrt{3}}{2}\) (C) \(\frac{1}{2}\) (D) \(\frac{1}{4}\)

Step-by-Step Solution

Answer

(C) \(\frac{1}{2}\)

Step 1: Understand the Problem

We are given two circles: Circle \(C\) with center \((1,1)\) and radius 1, and another circle \(T\) with center \((0, y)\) that passes through the origin and touches circle \(C\) externally. Our task is to find the radius of circle \(T\).

Step 2: Find the Radius of Circle T

Since \(T\) passes through the origin, the radius of \(T\) is equal to the distance from the center \((0, y)\) to the origin \((0, 0)\). Thus, radius \(r_T = y\).

Step 3: Use the External Tangency Condition

Circle \(T\) is externally tangent to circle \(C\). Therefore, the sum of their radii equals the distance between their centers. Denote the radius of \(C\) as \(r_C = 1\). The distance between the centers is given by the equation: \(\sqrt{(1-0)^2 + (1-y)^2} = 1 + r_T\).

Step 4: Solve the Equation for y

Substitute the distance formula into the tangency condition: \[ \sqrt{1 + (1-y)^2} = 1 + y \]Square both sides to eliminate the square root:\[ 1 + (1-y)^2 = (1 + y)^2 \]Expand both sides:\[ 1 + 1 - 2y + y^2 = 1 + 2y + y^2 \]This simplifies to:\[ 2 - 2y = 2y \]Solve for \(y\):\[ 2 = 4y \] \[ y = \frac{1}{2} \]

Step 5: Confirm the Radius of Circle T

Since we previously defined the radius of \(T\) as \(y\), the radius of circle \(T\) is \(\frac{1}{2}\).

Key Concepts

External Tangency

Two circles are said to touch each other externally when they meet at exactly one point. This occurs without either circle intersecting or covering the other. The point where they touch is called the point of tangency.
When dealing with external tangency, the rule is that the distance between the centers of the two circles is equal to the sum of their radii.
For example:

• If circle A has a radius of 3 and circle B a radius of 2, and they are externally tangent, then the distance between their centers must be 5, because 3 + 2 = 5.

Understanding external tangency is crucial in solving geometry problems involving circles, just like in the exercise described. By knowing how the distance between the centers relates to the radii, we can find unknown values effectively.

Distance Formula

The distance formula is a tool used to calculate the distance between two points on a coordinate plane. For two points \((x_1, y_1)\) and \((x_2, y_2)\), the formula to find the distance \(d\) is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
In context with circles and tangency, the distance formula helps to calculate the distance between the centers of two circles. This is especially important in establishing the condition for tangency.
Here’s a digestible example:

• Imagine two points, Point A at (3,4) and Point B at (6,8). By applying the distance formula: \[ \sqrt{(6-3)^2 + (8-4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
• This value is the straight-line distance between the two points.

Once you grasp the use of distance formulae, it becomes a powerful method in solving complex geometrical problems.

Solving Equations

Solving equations is a fundamental skill in mathematics that involves finding the values of variables that make the equation true. For a problem like the exercise, it allows us to find unknown quantities, such as the radius of a circle, by manipulating given equations.
Here's how you can approach solving equations:

• First, clearly understand what is given and what you need to find.
• Arrange the equations such that the unknowns are isolated on one side of the equation.
• Use algebraic operations such as addition, subtraction, multiplication, division, and, sometimes, more complex operations like squaring both sides to simplify the equation.

An example from the exercise involves:

• You have an equation involving square roots: \( \sqrt{1 + (1-y)^2} = 1 + y \)
• To solve, square both sides to get rid of the square root, simplify the equation, and then solve for \(y\).
• The steps involved require careful handling to ensure that the algebraic manipulations are correct, eventually leading to finding \( y = \frac{1}{2} \).

Mastering solving equations is about practice and understanding the logical flow of mathematics.