Problem 23 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Two circles with equal radii are intersecting at the points \((0,1)\) and \((0,-1)\). The tangent at the point \((0,1)\) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is : (a) 1 (b) 2 (c) \(2 \sqrt{2}\) (d) \(\sqrt{2}\)

Step-by-Step Solution

Answer

The distance between the centers is 2.

Step 1: Analyze the Given Information

We know that two circles with equal radii intersect at points \((0, 1)\) and \((0, -1)\). Let the centers of the circles be \((h_1, k_1)\) and \((h_2, k_2)\). We also know that the tangent at \((0, 1)\) to one of the circles passes through the center of the other circle.

Step 2: Understand the Tangent Condition

A tangent at any point of a circle is perpendicular to the radius at that point. Here, the tangent at \((0, 1)\) of one circle is passing through the center of another circle. This implies that the line joining the center of the second circle and point \((0, 1)\) is perpendicular to the line joining the center of the first circle to \((0, 1)\).

Step 3: Calculate Slope and Condition for Perpendicular Lines

The slope of the line joining the center of one circle \((h_1, k_1)\) to \((0, 1)\) is \(\frac{1 - k_1}{0 - h_1}\), and for the line joining \((h_2, k_2)\) to \((0, 1)\) is \(\frac{1 - k_2}{0 - h_2}\). For these lines to be perpendicular, the product of their slopes should be \(-1\). Therefore: \[\frac{1 - k_1}{-h_1} \times \frac{1 - k_2}{-h_2} = -1\]

Step 4: Use Symmetry of the Circle and Tangent

Since both circles intersect at the points \((0, 1)\) and \((0, -1)\), we can assume that symmetrically, the centers of the circles can be \((h, 0)\) and \((-h, 0)\). This is just one permissible configuration due to symmetry, simplifying our setup.

Step 5: Calculate the Distance Between Centers

With centers at \((h, 0)\) and \((-h, 0)\), the distance between these centers is \(2h\). Since the tangent at \((0, 1)\) should pass through the other center, this corresponds to radius \(\sqrt{h^2 + 1}\). For the tangent, we have the relation: \[-h^2 = -h^2\] This simplifies to \(h = 1\), deduced from initial conditions.

Step 6: Conclude the Distance Between Centers

Therefore, the distance between the centers is \(2h = 2 \times 1 = 2\).

Key Concepts

Circle Tangent

A tangent to a circle is a straight line that touches the circle at exactly one point. This line is perpendicular to the radius at the point of tangency. In our exercise, two circles intersect at points

• (0, 1)
• (0, -1)

It is given that a tangent at the point (0, 1) on one circle passes through the center of the other circle.
This bit of information is crucial, as it indicates a special arrangement; the tangent is not only at the intersection point but also exactly aligned to create a line through both the tangent point and the opposite circle’s center. Understanding how tangents function is key to solving problems involving circle geometry.
Tangents help us determine relationships between two circles, especially in cases where these relationships influence the angles and positions of points.

Symmetric Configuration

A symmetric configuration implies that two or more objects are arranged in a way that mirrors each other. This concept is often utilized in geometry to simplify problems, specifically concerning circles.
In our exercise, recognizing symmetry sheds insight. Because circles intersect at two specific points

• (0, 1)
• (0, -1)

we can assume symmetric placement for the centers of our circles.
By placing the centers at (h, 0) and (-h, 0), we leverage symmetry. This simplification reflects real properties of the circles, helping us better explore the geometric relationships, and leading us efficiently to measure necessary distances or resolve tangential conditions.
Easily solving geometric configurations often involve symmetric assumptions, reducing problem complexity.

Tangent Condition

The tangent condition is essential in understanding geometric layouts. It relates to how a tangent line's properties constrain and describe the circle it touches.
In this context, we identify two critical tangents:

• The tangent at (0, 1) of one circle passes through the center of the opposite circle.

This configuration means the radius to (0, 1), the tangent point, is perpendicular to a line from one center to the other. This defines a certain perpendicularity.
For our exercise, it comes down to applying slope rules:

• Compute slopes from circle centers to (0, 1) with the known relation
• \(rac{1 - k_1}{-h_1} \)
• \(rac{1 - k_2}{-h_2} \)

The perpendicularity implies their product equals -1, driving us to our eventual conclusion through straightforward calculations.

Distance Between Circle Centers

The distance between centers of two circles is key in determining how they relate spatially.
This exercise provides centers as (h, 0) and (-h, 0). Symmetry dictates this, simplifying calculation.
To find the distance, the formula;

• \(2h\)

applies because:

• Both centers are horizontally placed.

The distance stems from summarizing these features. Given symmetry, one arrives at a solution after reasoning that

• theses centers are equally spaced about the origin

. So, the distance resolves accurately to 2 when h = 1. Detailed understanding of symmetrical and tangential properties aids in deriving this conceptualization and applying proper geometric principles.