Problem 76 - A Complete Resource Book in Mathematics for JEE Main

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) \(16 x+8 y+25=0\) (B) \(6 x-8 y+25=0\) (C) \(8 x+6 y-25=0\) (D) \(6 x-8 y-25=0\)

Step-by-Step Solution

Answer

The correct options are B and D.

Step 1: Find the Length of the Diagonal

The formula for the length of the diagonal of a rectangle is the distance between its end points. Given two endpoints of the diagonal \((x_1, y_1)\) and \((x_2, y_2)\), the formula is \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Here, the endpoints are \((0,0)\) and \((4, 3)\). We have \( \text{Length of the diagonal} = \sqrt{(4 - 0)^2 + (3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5.\)

Step 2: Equation of Diagonal as a Line

The slope \(m\) of the line passing through points \((0,0)\) and \((4,3)\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{4 - 0} = \frac{3}{4}\). The equation of the line using point-slope form \(y - y_1 = m(x - x_1)\) becomes \(y = \frac{3}{4}x\). To eliminate fractions, multiply through by 4: \(4y = 3x\), which can be written as \(3x - 4y = 0\).

Step 3: Equation of Parallel Lines

Lines parallel to \(3x - 4y = 0\) have the same slope. Therefore, they are of the form \(3x - 4y = c\), where \(c\) is some constant. We need to find these lines which also touch the circumcircle of the rectangle.

Step 4: Finding the Circumcenter and Circumradius

The circumcenter of a rectangle is the midpoint of its diagonal, \((x,y) = \left(\frac{0 + 4}{2}, \frac{0 + 3}{2}\right) = (2, 1.5)\). The circumradius \(R\) is half the diagonal, so \(R = \frac{5}{2}.\)

Step 5: Use Point-Slope Distance Formula for Tangents

The distance from the line \(3x - 4y = c\) to the circumcenter \((2, 1.5)\) must be equal to the circumradius \(\frac{5}{2}\). Using the point-line distance formula \( \text{Distance} = \frac{|3(2) - 4(1.5) + c|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 6 + c|}{5} = \frac{|c|}{5}\). Equating to \(\frac{5}{2}\), we have \(|c| = \frac{5}{2} \times 5 = 12.5\).

Step 6: Final Equation of Tangents

Thus, \( c = \pm 12.5 \). Therefore, the two tangent equations are \(3x - 4y = 12.5\) and \(3x - 4y = -12.5\). Multiply through to eliminate fractions: \(6x - 8y = 25\) and \(6x - 8y = -25\). So, the correct equations are \(6x - 8y + 25 = 0\) and \(6x - 8y - 25 = 0\).

Key Concepts

Diagonals of a Rectangle

For any rectangle, the diagonals have a special property: they are equal in length. This means if you know one diagonal's length, you automatically know the other. The diagonals intersect at the midpoint, effectively dividing each other into two equal parts at 90 degrees. This forms the basis of many geometric calculations.

To calculate the length of a diagonal in a rectangle, you use the distance formula:

• Given endpoints ((x_1, y_1)) and ((x_2, y_2)), the diagonal length is (\( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)).

In this exercise, the endpoints ((0,0)) and ((4,3)) make use of this formula, resulting in a diagonal length of 5.

Remember, knowing the position of diagonals helps you understand other properties of the rectangle, like the possible dimensions or confirming right angles.

Circumcircle of a Rectangle

The circumcircle of a rectangle is a circle that passes through all four vertices of the rectangle. Every rectangle can be enclosed in its circumcircle, and this circle's center, known as the circumcenter, is at the intersection of the diagonals.

For any rectangle:

• The circumcenter is at the midpoint of any diagonal.
• The circumradius is half the length of a diagonal.

Using the rectangle in the given problem, the midpoint of the diagonal from ((0,0)) to ((4,3)) is ((2, 1.5)). This point is the circumcenter. The circumradius, being half the diagonal, measures (\(2.5\)).

This understanding aids in locating the tangents to the rectangle's circumcircle, which are parallel or perpendicular to its diagonals.

Distance Formula

The distance formula is a fundamental tool in geometry, allowing you to calculate the length between any two points in the coordinate plane. It is derived from the Pythagorean theorem and is expressed as:

• \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

This formula is critical for determining lengths of sides, diagonals, and radii. For instance, you determined the diagonal's length in the given problem using the distance formula. It also helps in finding equations of lines or circles by providing necessary distance metrics between points.

The exercise uses the point-line distance formula:

• \[ \text{Distance from line} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]

This formula helps find the distance from the circumcenter to the tangents of the circumcircle. Solving these using distances ensures you maintain exact alignments needed in geometry for accurate tangent construction.