Problem 61 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Extremities of a diagonal of a rectangle are \((0,0)\) and \((4,3)\). The equations of the tangents to the circumcircle of the rectangle which are parallel to this diagonal are (A) \(16 x+8 y \pm 25=0\) (B) \(6 x-8 y \pm 25=0\) (C) \(8 x+6 y \pm 25=0\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (B) \(6x - 8y \pm 25 = 0\).

Step 1: Determine the Slope of the Diagonal

Calculate the slope of the diagonal connecting the points \((0,0)\) and \((4,3)\). The formula for the slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For our points, the slope is \(m = \frac{3 - 0}{4 - 0} = \frac{3}{4}\).

Step 2: Find the Center and Radius of the Circumcircle

Since the rectangle has the diagonal endpoints at \((0,0)\) and \((4,3)\), its center is the midpoint of this diagonal, computed as \(\left( \frac{0+4}{2}, \frac{0+3}{2} \right) = (2, 1.5)\). The distance formula gives the radius: \(\sqrt{(2-0)^2 + (1.5-0)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5\).

Step 3: Write the Equation of Tangents Parallel to the Diagonal

The slope of the tangent lines parallel to the diagonal is also \(\frac{3}{4}\). Using the point-slope form of a line equation \(y - y_1 = m(x - x_1)\), where the line is parallel with the slope \(\frac{3}{4}\), we find equations through a set distance from the origin (radius of the circle) using the formula for the distance from a point to a line: \[ \frac{|3x - 4y + c|}{\sqrt{3^2 + 4^2}} = 2.5 \]. Here, \(\sqrt{3^2 + 4^2} = 5\).

Step 4: Solve for the Constants in the Line Equation

Rearrange the distance equation for \(|c|\) as follows: \[ \frac{|c|}{5} = 2.5 \implies |c| = 12.5 \]. So, the equations of the tangents are \(3x - 4y \pm 12.5 = 0\) or equivalently, simplify to \(6x - 8y \pm 25 = 0\) by multiplying through by 2.

Key Concepts

Understanding the Slope of a Diagonal

The slope of a diagonal in a rectangle is an essential concept when solving geometry problems involving lines and angles. The slope is like the measure of steepness of that diagonal and is calculated using the change in vertical position (the y-values) divided by the change in horizontal position (the x-values) between two points. For the diagonal of a rectangle joining points

• the coordinates i.e. (0, 0) and (4, 3), it's calculated as \( m = \frac{y_2 - y_1}{x_2 - x_1} \), where \( (x_1, y_1) = (0, 0) \) and \( (x_2, y_2) = (4, 3) \). The math here gives us a slope of \( m = \frac{3}{4} \).

This fraction means that as you move 4 units horizontally across (x-axis), you move 3 units vertically (y-axis). Understanding this ratio is crucial, as it helps us determine how other lines might be oriented relative to this diagonal. Lines that run parallel to this diagonal will have the same slope of \( \frac{3}{4} \), meaning they rise and run with the same ratio as this unique line.

Writing the Equation of a Line

The equation of a line is fundamental in conveying the relationship between x and y along any straight path. In geometry, the most common form of a line's equation is the slope-intercept form: \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept, or the point where the line crosses the y-axis.
But, when you're given only a slope and need to determine a line that is parallel to a known diagonal, you can instead use the point-slope form. This form emphasizes how the line's slope and a specific point determine its path. The point-slope equation is written as:

• \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a known point on the line and \( m \) is the slope the line shares with the diagonal.

Using our example's diagonal slope \( \frac{3}{4} \), you insert any specific point or multiple trials to figure out the point of tangency for tangent lines. This framework helps further determine how far the line shifts vertically (or horizontally) because it's parallel.

Finding the Distance from a Point to a Line

Understanding the distance from a point to a line involves a calculation that measures how far apart these entities are in the coordinate plane. This distance can be pivotal in various geometry scenarios like finding a tangent line

• , where precision is crucial.

The formula for the perpendicular distance from a point\((x_0, y_0)\) to a line\(Ax + By + C = 0\) is \[ \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \].
This sophisticated formula gives a clear direct measurement when you pop the values in. Relative to the circumcircle, you use this to ensure a line equidistant to the center can graze a curve or figure, like a circle, making it tangent. Given 5 (this being the norm formed from the diagonal)from the center, we set the left side of the earlier equation equal to 2.5 (corresponding to the proper radius length). Solving, it confirms when |c| goes as 12.5, finalizing the tangent’s mathematical set-up conforming to the line properties discussed.