6.1 Q-6

Question Statement

Show that the lines $3x - 2y = 0$ and $2x + 3y - 13 = 0$ are tangents to the circle given by the equation:

$$x^2 + y^2 + 6x - 4y = 0$$

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Background and Explanation

To prove that the given lines are tangents to the circle, we need to:

1. Find the center and radius of the circle.
2. Use the perpendicular distance formula to calculate the distance from the center of the circle to the lines.
3. Show that the perpendicular distances from the center to each of the lines are equal to the radius of the circle, which will confirm that the lines are indeed tangents.

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Solution

Step 1: Finding the Center and Radius of the Circle

The equation of the circle is given as:

$$x^2 + y^2 + 6x - 4y = 0 \tag{1}$$

To find the center and radius, we rewrite this equation in a more familiar form by completing the square for both $x$ and $y$.

• For $x$-terms: $x^2 + 6x$, add and subtract $\left(\frac{6}{2}\right)^2 = 9$.
• For $y$-terms: $y^2 - 4y$, add and subtract $\left(\frac{-4}{2}\right)^2 = 4$.

So, the equation becomes:

$$(x + 3)^2 + (y - 2)^2 = 13$$

From this, we can see that:

• The center of the circle is $C(-3, 2)$.
• The radius of the circle is $\sqrt{13}$.

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Step 2: Perpendicular Distance from the Center to the Line $3x - 2y = 0$

To find the perpendicular distance from the center $(-3, 2)$ to the line $3x - 2y = 0$, we use the formula for the perpendicular distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$:

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

For the line $3x - 2y = 0$, we have $A = 3$, $B = -2$, and $C = 0$. Substituting the values into the formula:

$$\text{Distance} = \frac{|3(-3) - 2(2) + 0|}{\sqrt{3^2 + (-2)^2}} = \frac{| -9 - 4 |}{\sqrt{9 + 4}} = \frac{13}{\sqrt{13}} = \sqrt{13} \tag{2}$$

Step 3: Perpendicular Distance from the Center to the Line $2x + 3y - 13 = 0$

Now, to find the perpendicular distance from the center $(-3, 2)$ to the line $2x + 3y - 13 = 0$, we again use the perpendicular distance formula.

For the line $2x + 3y - 13 = 0$, we have $A = 2$, $B = 3$, and $C = -13$. Substituting the values into the formula:

$$\text{Distance} = \frac{|2(-3) + 3(2) - 13|}{\sqrt{2^2 + 3^2}} = \frac{| -6 + 6 - 13 |}{\sqrt{4 + 9}} = \frac{13}{\sqrt{13}} = \sqrt{13} \tag{3}$$

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Step 4: Conclusion

Since the perpendicular distances from the center $C(-3, 2)$ to both lines $3x - 2y = 0$ and $2x + 3y - 13 = 0$ are both equal to the radius of the circle $\sqrt{13}$, we can conclude that both lines are tangents to the circle.

Thus, the lines $3x - 2y = 0$ and $2x + 3y - 13 = 0$ are tangents to the circle $x^2 + y^2 + 6x - 4y = 0$.

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Key Formulas or Methods Used

• Perpendicular Distance Formula:
  The perpendicular distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is:

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

• Equation of a Circle:
  To find the center and radius of the circle, rewrite the given equation in the standard form $(x - h)^2 + (y - k)^2 = r^2$ by completing the square.

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Summary of Steps

1. Step 1: Find the center and radius of the circle by rewriting the equation in standard form.
2. Step 2: Use the perpendicular distance formula to calculate the distance from the center to the line $3x - 2y = 0$.
3. Step 3: Use the perpendicular distance formula to calculate the distance from the center to the line $2x + 3y - 13 = 0$.
4. Step 4: Compare the distances with the radius and conclude that the lines are tangents to the circle.