6.3 Q-1

Question Statement

Prove that normal lines of a circle pass through the center of the circle.

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Background and Explanation

To solve this problem, we need to understand the basic concepts of tangent lines and normal lines in the context of circles. A tangent is a line that touches the circle at exactly one point and is perpendicular to the radius at that point. A normal line is a line that is perpendicular to the tangent line at the point of tangency. The key idea here is to prove that the normal line always passes through the center of the circle.

The equation of a circle centered at the origin is given by:

$$x^2 + y^2 = a^2$$

where $a$ is the radius of the circle.

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Solution

We are given the equation of a circle $x^2 + y^2 = a^2$, and we are tasked with proving that the normal line at any point on the circle passes through the center of the circle.

Step 1: Equation of the Tangent Line

Consider a point $(x_1, y_1)$ on the circle. The equation of the tangent line to the circle at this point is given by:

$$x x_1 + y y_1 - a^2 = 0$$

This is the equation of the tangent line because it satisfies the condition that the line touches the circle at exactly one point.

Step 2: Slope of the Tangent Line

The slope of the tangent line at the point $(x_1, y_1)$ can be found by differentiating the equation of the circle. For simplicity, we use the slope formula directly from the tangent equation. The slope of the tangent line is:

$$\text{Slope of tangent} = -\frac{x_1}{y_1}$$

Step 3: Slope of the Normal Line

The normal line to the circle at the point $(x_1, y_1)$ is perpendicular to the tangent line. The slope of a normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is:

$$\text{Slope of normal} = \frac{x_1}{y_1}$$

Step 4: Equation of the Normal Line

The normal line passes through the point $(x_1, y_1)$ and has a slope of $\frac{x_1}{y_1}$. Using the point-slope form of the equation of a line, we can write the equation of the normal line as:

$$y - y_1 = \frac{x_1}{y_1}(x - x_1)$$

Simplifying this equation:

$$y x_1 - y_1 x_1 = y_1(x - x_1)$$ $$y x_1 = y_1 x$$ $$y = \frac{y_1}{x_1} x$$

Step 5: Verifying the Normal Line Passes Through the Origin

Now, substitute the point $(0, 0)$ (the center of the circle) into the equation of the normal line:

$$0 = \frac{y_1}{x_1} \times 0$$

This equation holds true, which means that the normal line passes through the center of the circle.

Conclusion:

Since the equation of the normal line is satisfied by the center of the circle $(0, 0)$, we have proven that normal lines of a circle always pass through the center.

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Key Formulas or Methods Used

1. Equation of the Circle:

$$x^2 + y^2 = a^2$$

2. Slope of Tangent:

$$\text{Slope of tangent} = -\frac{x_1}{y_1}$$

3. Slope of Normal:

$$\text{Slope of normal} = \frac{x_1}{y_1}$$

4. Equation of the Normal Line:

$$y = \frac{y_1}{x_1} x$$

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Summary of Steps

1. Write the equation of the tangent line at a point $(x_1, y_1)$ on the circle.
2. Calculate the slope of the tangent line.
3. Determine the slope of the normal line (negative reciprocal of the tangent slope).
4. Derive the equation of the normal line using the point-slope form.
5. Verify that the normal line passes through the center of the circle by substituting $(0, 0)$ into the normal line equation.