Problem 67 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation of the chord of the circle \(x^{2}+y^{2}=a^{2}\) passing through the point \((2,3)\) and farthest from the centre is (A) \(2 x+3 y=13\) (B) \(3 x+2 y=13\) (C) \(2 x-3 y=13\) (D) none of these

Step-by-Step Solution

Answer

Option (A) \(2x + 3y = 13\) is correct.

Step 1: Understand the Problem

We are given the equation of a circle, \(x^2 + y^2 = a^2\), and we need to find the equation of a chord passing through a point \(2,3\) that is farthest from the center \((0,0)\).

Step 2: Determine the Perpendicular Distance from Point to Center

To find the chord of the circle that is farthest from the center, it should be perpendicular to the line joining the center of the circle \((0,0)\) to the given point \((2,3)\). The slope of the line from the center to \((2,3)\) is \(\frac{3-0}{2-0} = \frac{3}{2}\).

Step 3: Find the Slope of the Required Chord

The chord must be perpendicular to the line with slope \(\frac{3}{2}\), so the slope of the chord will be the negative reciprocal of \(\frac{3}{2}\), which is \(-\frac{2}{3}\).

Step 4: Write the Equation of the Chord

Using the point-slope form of the line equation, \(y - y_1 = m(x - x_1)\), with point \((2,3)\) and slope \(-\frac{2}{3}\), we write:\[y - 3 = -\frac{2}{3}(x - 2)\].

Step 5: Simplify the Equation

Rearranging the equation, we get:\[3(y - 3) = -2(x - 2)\]\[3y - 9 = -2x + 4\]\[2x + 3y = 13\].

Step 6: Verify the Solution

We check if this final result matches the given options. Option (A) is \(2x + 3y = 13\), which matches our equation.

Key Concepts

Understanding the Circle Equation

When you see an equation like \(x^2 + y^2 = a^2\), it represents a circle on a coordinate plane. Here, \(a\) is the radius of the circle and the center is at the origin \((0, 0)\). Every point \((x, y)\) that satisfies this equation is precisely \(a\) units away from the center. To visualize, imagine drawing all the points equidistant from the center; these points collectively form a perfect circle.

It's essential to recognize the standard form of a circle's equation, as it allows you to identify and plot the circle easily:

• Center: \((h, k)\) - in our case, \((0, 0)\).
• Radius: \(a\) - the distance from the center to any point on the circle.

This equation forms the basis for many problems involving circles, just like in our exercise where we need to find a specific chord.

Calculating the Slope of a Line

Understanding the slope of a line is crucial when dealing with geometry problems, especially those involving chords on a circle. The slope essentially measures the steepness and direction of a line. Given two points \((x_1, y_1)\) and \((x_2, y_2)\), the formula for the slope \(m\) is:\[m = \frac{y_2 - y_1}{x_2 - x_1}\]In our exercise, the slope connects the center of the circle \((0,0)\) to the point \((2,3)\).

Plugging in these values, we find:

• \((x_1, y_1) = (0,0)\)
• \((x_2, y_2) = (2,3)\)
• \(m = \frac{3-0}{2-0} = \frac{3}{2}\)

This slope is an important part of finding the line that is perpendicular to it, as seen in our solution.

Using the Point-Slope Form

The point-slope form of a line is an easy-to-use formula that allows you to find the equation of a line when you know the slope and a point on the line. The formula is:\[y - y_1 = m(x - x_1)\]Here, \((x_1, y_1)\) is the point, and \(m\) is the slope. In our exercise, the point used is \((2,3)\) and the slope needed is the negative reciprocal of \(\frac{3}{2}\), which is \(-\frac{2}{3}\).

Why the negative reciprocal? Because the line we want for the chord must be perpendicular to the original line.

• The perpendicular slope to \(\frac{3}{2}\) is \(-\frac{2}{3}\).
• Substituting into the point-slope form gives us:\[y - 3 = -\frac{2}{3}(x - 2)\]

This form is incredibly helpful, especially when constructing line equations quickly.

Understanding Perpendicular Lines

In geometry, a perpendicular line means two lines intersect at a right angle (90 degrees). On a coordinate plane, for two lines to be perpendicular, the product of their slopes must equal \(-1\). So, if one line has a slope of \(m\), then a line perpendicular to it will have a slope of \(-\frac{1}{m}\).

This is precisely the problem we tackled in our exercise. We needed to find a line perpendicular to another that had a slope of \(\frac{3}{2}\). By using the negative reciprocal, \(-\frac{2}{3}\), we ensured the two lines would meet at a right angle. This property is crucial for solving many geometric problems involving shapes and angles.

• Original Line Slope: \(\frac{3}{2}\)
• Perpendicular Line Slope: \(-\frac{2}{3}\)

Recognizing and utilizing this relationship between slopes helps determine when lines are at right angles to each other, like when determining chords on a circle.