6.2 Q-7

Question Statement

Find the equation of the tangent to the circle $x^2 + y^2 = 2$ that is:

1. Parallel to the line $x - 2y + 1 = 0$
2. Perpendicular to the line $3x + 2y = 6$

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Background and Explanation

To solve this problem, we need to use the properties of tangents to a circle. Specifically, we apply the following principles:

1. Parallel Tangents: For a line to be parallel to a given line, it must have the same slope. The equation of the tangent can be written in the general form of the line equation, and we use the condition that the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

2. Perpendicular Tangents: If a line is perpendicular to a given line, its slope is the negative reciprocal of the slope of the given line. We use the same condition as above: the perpendicular distance from the center of the circle to the tangent must be equal to the radius of the circle.

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Solution

i. Tangent Parallel to the Line $x - 2y + 1 = 0$

1. Equation of Parallel Tangent: The equation of any line parallel to $x - 2y + 1 = 0$ is of the form:

$$x - 2y + k = 0 \tag{1}$$

2. Condition for Tangency: For the line $x - 2y + k = 0$ to be tangent to the circle $x^2 + y^2 = 2$, the perpendicular distance from the center of the circle $(0, 0)$ to the line must be equal to the radius of the circle. The formula for the perpendicular distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is:

$$\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Substitute the center $(0, 0)$ and the line $x - 2y + k = 0$:

$$\frac{|0 - 2(0) + k|}{\sqrt{1^2 + 2^2}} = \sqrt{2}$$

This simplifies to:

$$\frac{|k|}{\sqrt{5}} = \sqrt{2}$$

3. Solve for $k$:

$$|k| = \sqrt{2} \times \sqrt{5} = \sqrt{10}$$

Thus, $k = \pm \sqrt{10}$.

4. Equation of the Tangents: Therefore, the equation of the tangents parallel to $x - 2y + 1 = 0$ is:

$$x - 2y \pm \sqrt{10} = 0$$

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ii. Tangent Perpendicular to the Line $3x + 2y = 6$

1. Slope of the Given Line: The slope of the line $3x + 2y = 6$ is:

$$\text{Slope of given line} = \frac{-3}{2}$$

Since the tangent is perpendicular to this line, its slope will be the negative reciprocal:

$$\text{Slope of tangent} = \frac{2}{3}$$

2. Equation of the Tangent: The general equation of the tangent to the circle $x^2 + y^2 = 2$ with slope $\frac{2}{3}$ is:

$$y = \frac{2}{3}x \pm \sqrt{2} \sqrt{1 + \frac{4}{9}}$$

3. Simplify the Expression:

$$y = \frac{2}{3}x \pm \sqrt{2} \times \frac{\sqrt{13}}{3}$$ $$y = \frac{2}{3}x \pm \frac{\sqrt{26}}{3}$$

4. Multiply through by 3 to simplify further:

$$3y = 2x \pm \sqrt{26}$$

Thus, the equations of the tangents perpendicular to $3x + 2y = 6$ are:

$$2x - 3y + \sqrt{26} = 0$$

and

$$2x - 3y - \sqrt{26} = 0$$

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Key Formulas or Methods Used

• Perpendicular Distance from a Point to a Line:

$$\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

• Equation of a Line Parallel to a Given Line: The equation $x - 2y + k = 0$ represents any line parallel to $x - 2y + 1 = 0$, where $k$ is a constant.
• Slope of Perpendicular Lines: The slope of the tangent line is the negative reciprocal of the slope of the given line.

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Summary of Steps

1. For Tangent Parallel to $x - 2y + 1 = 0$:

   • Set up the equation of the parallel line.
   • Use the condition that the perpendicular distance from the center to the tangent is equal to the radius of the circle.
   • Solve for $k$ to find the equation of the tangents.

2. For Tangent Perpendicular to $3x + 2y = 6$:

   • Find the slope of the given line and take the negative reciprocal for the tangent’s slope.
   • Use the general equation for the tangent line with the calculated slope.
   • Solve for the equation of the tangents.

The equations of the tangents are:

• Parallel: $x - 2y \pm \sqrt{10} = 0$
• Perpendicular: $2x - 3y \pm \sqrt{26} = 0$