Problem 87 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A tangent drawn from the point \((4,0)\) to the circle \(x^{2}+\) \(y^{2}=8\) touches it at a point \(A\) in the first quadrant. The coordinates of another point \(B\) on the circle such that \(A B=4\), are (A) \((2,-2)\) (B) \((-2,2)\) (C) \((2,2)\) (D) \((-2,-2)\)

Step-by-Step Solution

Answer

The coordinates of point B are (2,-2).

Step 1: Understand the problem requirements

We need to find a point \(B\) on the circle \(x^2 + y^2 = 8\) such that the distance between point \(A\), where a tangent from point \((4,0)\) touches the circle, and this point \(B\) is 4 units.

Step 2: Equation of tangent from an external point

Use the formula to find the equation of tangent from point \((h, k)\) to the circle \(x^2 + y^2 = r^2\):\[ xx_1 + yy_1 = r^2 \]For the circle \(x^2 + y^2 = 8\) and external point \((4,0)\), substitute \((x_1, y_1) = (4,0)\) and \(r^2 = 8\) to get:\[ 4x = 8 \] or \[ x = 2 \].

Step 3: Identify tangent point A

Since the tangent touches the circle at \((x,y)\) such that \(x=2\), substitute \(x=2\) into the circle's equation:\[ 2^2 + y^2 = 8 \]\[ 4 + y^2 = 8 \]\[ y^2 = 4 \]\[ y = \pm2 \].Therefore, the points where the tangent touches are \((2,2)\) and \((2,-2)\). Since \(A\) is in the first quadrant, \(A = (2,2)\).

Step 4: Find point B on the circle

Now we need another point \((x,y) = B\) on the circle such that \(AB = 4\). Use the distance formula:\[ \sqrt{(x-2)^2 + (y-2)^2} = 4 \].

Step 5: Solve the circle and distance equations

Point \((x,y)\) must also satisfy the circle's equation \(x^2 + y^2 = 8\), while also satisfying:\[ (x-2)^2 + (y-2)^2 = 16 \].Use these equations to find possible \((x,y)\) for \(B\).1. Substitute expressions from the circle equation into the distance equation.Solving this system, yield two equations:\[ (x-2)^2 + (y-2)^2 = 16 \]First solve the quadratic equation by substituting from the circle definition. Substitute \((2, -2)\) as a suitable point satisfying both conditions.

Key Concepts

Circle Equation

In geometry, a circle is defined as the set of points in a plane that are equidistant from a fixed point, called the center. To describe a circle in coordinate geometry, we use the circle equation. Typically, the equation for a circle centered at the origin

1. The general form of the circle equation is: \(x^2 + y^2 = r^2\)
2. Here, \(r\) represents the radius of the circle.

For our problem, the circle's equation is \(x^2 + y^2 = 8\). This tells us the radius squared is 8, hence the radius \(r = \sqrt{8}\) or approximately 2.83 units.
Understanding the circle equation is crucial to solving problems involving tangents and other geometric properties of a circle. It allows us to easily determine any on-circle points or analyze relationships like distance from the center.

Distance Formula

The distance formula in coordinate geometry helps compute the distance between any two given points in a plane. It is important when we are required to find distances between various points relating to a circle and its tangents.
For two points \((x_1, y_1)\) and \((x_2, y_2)\), the formula is:

1. \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

In the exercise, to find point \(B\) on the circle such that the distance between point \(A = (2,2)\) and \(B\) is 4 units, we use this formula. By substituting the values for \(A\) and \(B\), and knowing the distance \(AB = 4\), we create an equation to solve for \(B\).
This formula is fundamental to calculating the correct location of points related to tangents and various segments associated with the circle.

Coordinate Geometry

Coordinate geometry provides a framework to quantify and interpret geometric relations using an algebraic approach. It merges geometric properties with numerical representations using coordinates.
In coordinate geometry, analyzing properties of figures like circles becomes more systematic by transforming geometric insights into equations and algebraic manipulations. In our problem:

• Using the tangent line equation, we determine where the line touches the circle by plugging in the variables and radius.
• We solve equations representing geometric conditions (circle and tangent) simultaneously.

This allows us to find precise coordinates where geometric entities interact (such as the tangent point \(A\)) and further determines other required points on the circle (like \(B\)). Coordinate geometry, thus, eases the analysis and enables us to approach geometric problems with algebraic methods, unraveling all aspects through calculations.