Problem 75 - A Complete Resource Book in Mathematics for JEE Main

Question

The coordinates of a point on the line \(y=2\) from which the tangents drawn to the circle \(x^{2}+y^{2}=25\) are perpendicular, are (A) \((\sqrt{46}, 2)\) (B) \((-\sqrt{46}, 2)\) (C) \((\sqrt{37}, 2)\) (D) \((-\sqrt{37}, 2)\)

Step-by-Step Solution

Answer

Correct points are (A) \((\sqrt{46}, 2)\) and (B) \((-\sqrt{46}, 2)\).

Step 1: Understand the Given Line and Circle Equations

The line equation given is \( y = 2 \), which means all points on this line have a y-coordinate of 2. The circle equation given is \( x^2 + y^2 = 25 \), a circle centered at the origin with radius 5.

Step 2: Set Up Conditions for Perpendicular Tangents

The condition states that the tangents from the point \((a, 2)\) to the circle are perpendicular. Two tangents from a point \((x_1, y_1)\) to a circle \((x - h)^2 + (y - k)^2 = r^2\) are perpendicular if \(x_1h + y_1k = r^2\).

Step 3: Substitute and Solve for Coordinate

Substitute \(x_1 = a\), \(y_1 = 2\), \(h = 0\), \(k = 0\), and \(r = 5\) into the perpendicularity condition: \[ a \times 0 + 2 \times 0 = 25 \] Since this doesn't provide information, we compute \(a\) directly using the tangent line equation.

Step 4: Calculate the Length of Tangents

The length of the tangent from a point \((x_1, y_1)\) to a circle \(x^2 + y^2 = r^2\) is \(\sqrt{x_1^2 + y_1^2 - r^2}\). Substitute \((a, 2)\) and circle's radius, \(5\): \[ \sqrt{a^2 + 4 - 25} = \sqrt{a^2 - 21} = 5 \] Solve \( a^2 - 21 = 25 \) to find \( a^2 = 46 \).

Step 5: Determine Possible Values for Coordinates

Since \( a^2 = 46 \), solving gives us \( a = \pm \sqrt{46} \). Therefore, the possible points are \((\pm \sqrt{46}, 2)\).

Step 6: Identify Correct Options from Given Choices

From the possible coordinates, the given options are checked. The correct answers based on calculations are (A) \((\sqrt{46}, 2)\) and (B) \((-\sqrt{46}, 2)\).

Key Concepts

Circle Equation

A circle equation is a mathematical way to represent all the points in a plane that are equidistant from a single point, known as the center. The general equation of a circle with a center at

• the origin (0,0) is \( x^2 + y^2 = r^2 \).
• Here, \( r \) represents the radius of the circle.

This equation shows that for any point

• \((x, y)\) on the circle, the distance from the center is a constant value \( r \).

For example, the circle equation \( x^2 + y^2 = 25 \) implies that every point on the circle is 5 units away from the origin (since \( \sqrt{25} = 5 \)). This makes solving problems involving circles and tangents a systematic process that relies on these geometric properties.

Tangent Line

A tangent to a circle is a line that touches the circle at exactly one point. It is perpendicular to the radius drawn at the point of contact. This key characteristic helps solve many problems in geometry related to circles. To determine if a line

• is tangent to a circle, we use the slope properties and intersections.

The equation of a tangent line can be derived using the point \

• outside the circle from which the tangent is drawn and its corresponding point on the circle.

In the exercise, we use the line equation \( y = 2 \) to understand how tangents can be drawn from a point \((a, 2)\) and satisfy specific geometric conditions, making them extremely useful in calculations and prediction of behavior of such lines.

Perpendicular Tangents

When two tangents to a circle are perpendicular to each other, there is a special condition that links their geometry. For tangents drawn from a point \((x_1, y_1)\) to a circle centered at

• the origin (0,0) with radius \( r \), they are perpendicular if \( x_1h + y_1k = r^2 \) holds true.

In our specific example,

• the condition simplifies well due to the circle's center being the origin and the linear y-coordinate being constant.

This simplification turns the equation into a more accessible problem and helps us understand the nature of evaluating perpendicularity in the context of coordinate geometry.

Coordinate Calculation

Calculating the exact coordinates for tangents involves algebraic manipulation and understanding of the geometry involved in circle and tangent line interactions. In the case of perpendicular

• tangents, you will usually need to find the point of tangency that satisfies given conditions on slopes and perpendicularity.
• Given the equation of the circle \( x^2 + y^2 = 25 \) in our exercise, the task required finding a point \((a, 2)\).

We used the tangent length formula, \( \sqrt{a^2 + b^2 - r^2} = 5 \), to backtrack to possible values of \( a \).

• Solving \( a^2 = 46 \) gives the coordinates of the points from which the tangents drawn are \( \pm \sqrt{46} \), providing a comprehension of how algebra links with geometry in coordinate systems effectively.