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

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

The circle in option B is normal to the line and goes through the point.

Step 1: Understand Normal Line Definition

A line is normal to a circle at a given point if it passes through the center of the circle and the point itself. Since the given line equation is \(y - x + 3 = 0\), rearrange it to understand its slope: \(y = x - 3\). This line has a slope of 1.

Step 2: Determine Slope of Radius

The radius of the circle will be perpendicular to the tangent at the point \(\left(3+\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right)\). Since the given line is normal to the circle, its slope is also the slope of the radial line. Hence the slope of the radius must be -1, since it must be perpendicular to the tangent (slope 1).

Step 3: Define Center Using the Perpendicular Line

Since the radial line has a slope of -1, use the point \(\left(3+\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right)\) to write its equation: \(y - \frac{3}{\sqrt{2}} = -1(x - \left(3+\frac{3}{\sqrt{2}}\right)\). Simplify to find: \(y = -x + 3 + \frac{3}{2\sqrt{2}} + \frac{3}{\sqrt{2}} \). For a point on the radial line, \(x = \frac{3}{\sqrt{2}}\), \(y = \frac{3}{\sqrt{2}}\). Thus, the center of the circle is \(\left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right)\).

Step 4: Compare Center with Options

Now compare the calculated center \(\left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right)\) with the centers given in options A, B, C, and D. Only option B has a center \(\left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right)\) and matches the circle \(\left(x - \frac{3}{\sqrt{2}}\right)^2 + \left(y - \frac{3}{\sqrt{2}}\right)^2 = 9\).

Key Concepts

Equation of a Circle

The equation of a circle is a crucial algebraic expression representing all the points on a plane that maintain a consistent distance, known as the radius, from a fixed point called the center. The standard form of a circle's equation is given as \[(x - h)^2 + (y - k)^2 = r^2\]where:

• \(h, k\) are the coordinates of the circle's center.
• \(r\) denotes the radius of the circle.

This format helps quickly identify the circle's essential features simply by looking at the equation. In problems involving normal lines, determining the circle's center allows us to understand the circle's geometric context in relation to other lines or curves. Keep in mind how the center and radius relate to the point where a line is normal to a circle, as the line passes through the circle's center.

Geometric Interpretation of Slope

The slope of a line gives insight into its steepness and direction. Mathematically, slope is defined as the "rise over run," or the ratio of the vertical change to the horizontal change between two points on a line.\[m = \frac{\Delta y}{\Delta x}\]In a geometric context, the slope plays a fundamental role in defining the orientation of lines. For example, lines with a positive slope rise to the right, while those with a negative slope fall to the right.
When examining a line in relation to a circle, the slope tells us how the line interacts with the circle's curvature. For instance, if a radial line emanates from the center to a point on the circle where the slope is known, it helps in visualizing the line’s perpendicularity to tangent lines at that point, leading to the concept of normal lines.

Perpendicular Lines

Perpendicular lines intersect at a right angle, which is 90 degrees. A crucial feature of perpendicular lines in geometry is their slopes' relation. If two lines are perpendicular, the product of their slopes equals \(-1\) (assuming neither line is vertical, where slope would be undefined). For this reason, if you have one line with a slope \(m\), a perpendicular line will have a slope of \[-\frac{1}{m}\].Understanding this relationship is essential when dealing with circles, especially concerning normal lines. If a line is normal (perpendicular) to the circle at a point, it typically shares this perpendicular slope relation with the circle's radius at that point of contact.

• This characteristic allows us to find the circle's center, as the normal line must pass through this center, forming a right angle with the radius.
• Checking perpendicularity helps define geometric boundaries between linear and curved lines in exercises.