Problem 52 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(C\) be the circle with centre at \((1,1)\) and radius \(=1\). If \(T\) is the circle centred at \((0, y)\), passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal to (a) \(\frac{1}{2}\) (b) \(\frac{1}{4}\) (c) \(\frac{\sqrt{3}}{\sqrt{2}}\) (d) \(\frac{\sqrt{3}}{2}\)

Step-by-Step Solution

Answer

The radius of circle \(T\) is \(\frac{1}{4}\) (option b).

Step 1: Define the Centers and Radii

The circle \(C\) has its center at \((1,1)\) and radius \(r_C = 1\). The circle \(T\) is centered at \((0, y)\) and passes through the origin, so its radius \(r_T\) is equal to the distance from \((0, y)\) to \((0,0)\). Thus, \(r_T = y\).

Step 2: Use the Condition of External Tangency

Since circle \(T\) touches circle \(C\) externally, the distance between their centers equals the sum of their radii. Therefore, \(\sqrt{(1-0)^2 + (1-y)^2} = 1 + y\).

Step 3: Calculate the Distance Between Centers

Calculate the distance between centers: \[ \sqrt{1^2 + (1-y)^2} = 1 + y \]Simplifying the inside of the square root gives: \[ \sqrt{1 + (1 - 2y + y^2)} = \sqrt{2 - 2y + y^2} = 1 + y \].

Step 4: Square Both Sides to Eliminate the Square Root

Squaring both sides results in: \[ 2 - 2y + y^2 = (1 + y)^2 \]which simplifies to \[ 2 - 2y + y^2 = 1 + 2y + y^2 \].

Step 5: Simplify and Solve for y

Cancel \(y^2\) from both sides and rearrange terms:\[ 2 - 2y = 1 + 2y \]This simplifies to: \[ 2 - 1 = 2y + 2y \]\[ 1 = 4y \]Hence, \(y = \frac{1}{4}\).

Step 6: Conclusion

The radius of circle \(T\), which is equal to \(y\), is therefore \(\frac{1}{4}\).

Key Concepts

Circle Properties

In coordinate geometry, understanding the properties of a circle is crucial. A circle is defined by its center and radius. The center is a fixed point on the plane, and the radius is the distance from this center to any point on the circle. For example, for circle \(C\) centered at \((1,1)\) with a radius of 1, every point on circle \(C\) is exactly 1 unit away from the center point \((1,1)\).
A circle's equation can be written in the form \((x-h)^2 + (y-k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius. This equation states that for any point \((x,y)\) on the circle, the distance to the center \((h,k)\) is precisely \(r\).

• The set of all points that are equidistant from a central point forms a circle.
• The fixed distance is the radius, and the central point is the center.

Understanding these foundational properties helps when analyzing the relationships between different circles and when working with more complex scenarios like tangency.

External Tangency Condition

Tangency describes how circles interact or connect with each other on the plane. For two circles to touch externally, there is a specific condition that must be met. This condition states that the distance between their centers must equal the sum of their radii.
In the problem, circle \(T\), with its center at \((0,y)\), touches circle \(C\) externally. According to the external tangency condition:

• The formula to remember is \(\text{distance between centers} = r_1 + r_2\).
• With this, you can establish an equation lined up with the circle equations you have.

In this case, the distance between the center of circle \(C\) \((1,1)\) and the center of circle \(T\) \((0,y)\) is equal to \(1 + y\), where 1 is the radius of circle \(C\) and \(y\) is the radius of circle \(T\). By understanding this external tangency condition, we can set up equations and solve for unknown values such as the radius of circle \(T\).

Distance Formula

The distance formula is a fundamental tool in coordinate geometry that allows us to calculate the distance between two points on a plane. For two points, \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula helps us compute the exact separation between the centers of two circles, which is vital for determining tangency conditions or other interactions.

• For circle \(C\) with center \((1, 1)\) and circle \(T\) at \((0, y)\), the distance is calculated as \(\sqrt{(1-0)^2 + (1-y)^2}\).
• This simplifies using arithmetic operations, leading us to \(\sqrt{2 - 2y + y^2}\).

Understanding how to apply the distance formula effectively allows us to translate geometric scenarios into solvable equations, bridging the gap between visual and numeric interpretation.

Equation Solving

Equation solving is a process used to find unknown values by using algebraic manipulation. Once you set up equations based on geometric conditions, like the distance between centers equaling the sum of radii, your next step is to solve for any unknowns. Let's look at how this was done in our exercise.
You start with the equation:
\[ \sqrt{2 - 2y + y^2} = 1 + y \]
By squaring both sides, you eliminate the square root:
\[ 2 - 2y + y^2 = (1 + y)^2 \]
This process creates a standard algebraic equation, which can then be simplified and rearranged:

• Expand and simplify both sides: \( 2 - 2y + y^2 = 1 + 2y + y^2 \).
• Cancel terms and solve for \(y\): \( 2 - 2y = 1 + 2y \).
• Combine like terms to find \(y = \frac{1}{4}\).

Equations can often look intimidating, but breaking them down step by step reveals clear solutions. This approach allows you to navigate from complex real-world geometric scenarios to precise numeric answers.