Problem 9 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let the tangents drawn from the origin to the circle, \(x^{2}+y^{2}-8 x-4 y+16=0\) touch it at the points \(A\) and \(B\). The \(\begin{array}{ll}(A B)^{2} \text { is equal to: (a) \)\frac{52}{5}\( (b) \)\frac{56}{5}\( (c) \)\frac{64}{5}\( (d) \)\frac{32}{5}$

Step-by-Step Solution

Answer

The answer is (c) \(\frac{64}{5}\).

Step 1: Identify Circle Properties

The given equation of the circle is \(x^{2} + y^{2} - 8x - 4y + 16 = 0\). We need to rewrite this equation in its standard form \((x-h)^{2} + (y-k)^{2} = r^{2}\). Start by completing the square for both \(x\) and \(y\).

Step 2: Complete the Square

Rearrange and complete the square:- For \(x\), \((x^2 - 8x)\) becomes \((x - 4)^2 - 16\).- For \(y\), \((y^2 - 4y)\) becomes \((y - 2)^2 - 4\).Substitute back to get: \((x - 4)^2 + (y - 2)^2 - 16 - 4 + 16 = 0\).

Step 3: Simplify to Standard Circle Equation

The equation simplifies to \((x - 4)^2 + (y - 2)^2 = 4\). This is the standard form of a circle with center \((4, 2)\) and radius \(r = 2\).

Step 4: Determine the Tangent Distance Formula

The tangents from a point \((x_0, y_0)\) to a circle \((x-h)^2 + (y-k)^2 = r^2\) are given by the formula \(d = \sqrt{(x_0-h)^2 + (y_0-k)^2 - r^2}\).

Step 5: Calculate Distance from Origin to Circle's Center

Using the center of the circle \((4, 2)\):\[d = \sqrt{(4-0)^2 + (2-0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\].

Step 6: Use Tangent Length Formula

The length of each tangent from the origin is \(\sqrt{a^2 - r^2}\) where \(a = 2\sqrt{5}\) and \(r = 2\):\[\text{Tangent Length} = \sqrt{(2\sqrt{5})^2 - 2^2} = \sqrt{20 - 4} = \sqrt{16} = 4\].

Step 7: Calculate \((AB)^2\)

Since the tangents form a pair of tangent segments from a common external point, the length of line segment \(AB\) can be given by \(AB = 2 \times \text{Tangent Length} = 2 \times 4 = 8\). Thus, \((AB)^{2} = 8^2 = 64\).

Step 8: Determine Correct Answer from Options

Divide \((AB)^2\) obtained by 5 to match the options:\[\frac{64}{5}\]. Hence, the correct option is (c) \(\frac{64}{5}\).

Key Concepts

Circle Equation

A circle's equation in its standard form makes it easy to identify the circle's center and radius. The standard form looks like

• \((x-h)^2 + (y-k)^2 = r^2\)

where

• \((h, k)\) is the center of the circle
• \(r\) is the radius

To convert a general form equation to this standard form, we often need to simplify it using a method called completing the square. This will help us to precisely understand the circle's dimensions and its position on a coordinate grid. Such transformations are vital when dealing with tangents and their properties in mathematical problems.
Let's use these conversions to help analyze how points, like where tangents touch the circle, interact with the circle.

Completing the Square

Completing the square is a method used to convert a quadratic expression into a perfect square trinomial. This is very useful in identifying the features of a circle's equation.

• For the expression \(x^2 - 8x\), you rearrange it as \((x-4)^2 - 16\). This forms a perfect square, letting us recognize the component \((x-4)^2\).
• Similarly, for \(y^2 - 4y\), it becomes \((y-2)^2 - 4\).

By substituting these back into the circle's equation, we transform it into its standard form.The process highlights the center and radius of the circle clearly. In practice, being able to complete the square efficiently is key to tackling many geometry problems involving parabolas and circles.

Distance Formula

The distance formula is a pivotal tool for solving geometric problems. To find the distance between two points, such as a point and a circle's center, we use:

• \(d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\)

Here, \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.
This formula allows us to calculate the distance from the origin (0,0) to the circle's center, which is crucial for determining tangent properties.
In the exercise, this is precisely how the distance \(2\sqrt{5}\) is calculated from the origin to the center

• \((4, 2)\)

of the circle.

Length of Tangent

The length of a tangent from an external point to a circle is determined using the formula:

• \(\sqrt{a^2 - r^2}\)

where \(a\) is the distance from the external point to the circle's center, and \(r\) is the radius of the circle. This formula provides the length of one tangent.

• In our given problem, \(a = 2\sqrt{5}\) and \(r = 2\).

Subtract the square of the radius from the square of the distance, then find the square root to get the tangent length.
Knowing this length, you can solve for other properties related to the circle, such as finding the length of the tangent segments \(AB\) formed from common external points, showcasing how tangents relate to each other around a circular shape.