Problem 39 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

To which of the following circles, the line \(y-x+3=\) 0 is normal at the point \(\left(3+\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) ?\) (A) \(\left(x-3-\frac{3}{\sqrt{2}}\right)^{2}+\left(y-\frac{3}{\sqrt{2}}\right)^{2}=9\) (B) \(\left(x-\frac{3}{\sqrt{2}}\right)^{2}+\left(y-\frac{3}{\sqrt{2}}\right)^{2}=9\) (C) \(x^{2}+(y-3)^{2}=9\) (D) \((x-3)^{2}+y^{2}=9\)

Step-by-Step Solution

Answer

Option (A).

Step 1: Find Slope of Given Line

The equation of the line is given by \( y - x + 3 = 0 \). Rewriting it in the slope-intercept form, we have \( y = x - 3 \), indicating that the slope of this line is 1.

Step 2: Determine Slope of Normal Line

The normal line would have a slope that is the negative reciprocal of the given line's slope. Thus, the slope of the normal line is \(-1\).

Step 3: Identify the Equation of the Normal Line

The point given is \( \left(3+\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) \). The equation of the line normal at this point, with slope \(-1\), can be written using the formula \( y - y_1 = m(x - x_1) \): \[ y - \frac{3}{\sqrt{2}} = -1\left(x - \left(3 + \frac{3}{\sqrt{2}}\right)\right) \]. Simplifying, we get: \( y = -x + \left(3 + 2\frac{3}{\sqrt{2}}\right) \).

Step 4: Check the Circles

The center of each circle from the options should lie on the normal line: - A: Center \( \left(3+\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) \) lies on the line \( y = -x + \left(3 + 2\frac{3}{\sqrt{2}}\right) \).- B: Center \( \left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) \) does not satisfy the line's equation.- C: Center \( (0, 3) \) does not satisfy the line.- D: Center \( (3, 0) \) does not satisfy the line.

Step 5: Conclusion

Only the center in option (A) satisfies the equation of the normal. Thus, the line is normal to the circle given in option (A).

Key Concepts

Slope of a Line

The slope of a line is a way to describe how steep the line is. It tells us how much the line inclines or declines as we move from one point to another. Mathematically, if we have two points on a line, \( (x_1, y_1) \) and \( (x_2, y_2) \), then the slope can be calculated by the formula: \[m = \frac{y_2 - y_1}{x_2 - x_1}.\] For a line expressed in the standard form \( Ax + By + C = 0 \), you can rearrange it to the slope-intercept form \( y = mx + b \) to find the slope \( m \). In our example, the line was \( y - x + 3 = 0 \) which rearranges to \( y = x - 3 \), indicating the slope is 1. This means for each unit increase in \( x \), \( y \) increases by the same amount.

Equation of a Line

The equation of a line gives us a way to represent all points on the line in a simple equation. The most common form is the slope-intercept form: \( y = mx + b \). Here, \( m \) stands for the slope and \( b \) is the y-intercept—the point where the line crosses the y-axis. This form makes it easy to understand how the line is behaving on a graph.
In the exercise, we had a line with the equation \( y = x - 3 \), which tells us:

• The slope \( m \) is 1.
• The line crosses the y-axis at \( b = -3 \).

Another useful form is the point-slope formula: \( y - y_1 = m(x - x_1) \), which uses a known point \( (x_1, y_1) \) on the line. This is helpful when you have a specific point and the slope and want the equation of the line.

Normal Line

A normal line is a line perpendicular to a tangent line at a given point on a curve. In simple words, it is the line that forms a right angle with the tangent line. Calculating the slope of a normal line involves taking the negative reciprocal of the tangent line's slope.
For instance, if the slope of a given line is 1, then the slope of the normal line would be \( -1 \), since \[ m_{\text{normal}} = -\frac{1}{m_{\text{original}}}. \] In the exercise solution, we used this rule to find that the normal line to the given line at a specific point had a slope of \(-1\). This idea is vital in many mathematical fields, as normal lines can often represent the shortest path to a curve from a given point.

Center of a Circle

The center of a circle is a point within it, equidistant from all points on the circle itself. The center is represented in the general circle equation: \((x - h)^2 + (y - k)^2 = r^2\), where \( (h, k) \) denotes the center and \( r \) is the radius of the circle.
In the exercise, knowing where the center is helped determine which line is normal to the given circles.
Having the center on a normal line means \( (h, k) \) must satisfy the equation of the normal line.

• The given line was \( y = -x + \left(3 + 2\frac{3}{\sqrt{2}}\right) \).
• Checking centers allowed us to find that only the circle in option (A) matched.

Understanding this concept is essential when geometry problems involve circles and lines because it gives a definite point in space from which distances can be measured.