6.2 Q-8

Question Statement

Find the equations of the tangents drawn from the following points to the given circles:

1. From $(0, 5)$ to the circle $x^2 + y^2 = 16$.
2. From $(-1, 2)$ to the circle $x^2 + y^2 + 4x + 2y = 0$.
3. From $(x+1)^2 + (y-2)^2 = 26$.

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Background and Explanation

To find the equations of the tangents to a circle from an external point, we use a few key geometric principles:

1. General Form of the Tangent: The equation of a tangent to a circle is given by $y = mx + c$, where $m$ is the slope of the tangent and $c$ is the y-intercept.
2. Condition for Tangency: For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.
3. Use of Pythagorean Theorem: The length of the tangent from a point to the circle is related to the distance from the point to the center of the circle.

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Solution

i. Tangents from $(0, 5)$ to the Circle $x^2 + y^2 = 16$

1. Equation of the Tangent: The equation of any tangent to a circle is $y = mx + c$, where $m$ is the slope, and $c$ is the y-intercept. For the circle $x^2 + y^2 = 16$, the general form of the tangent is:

$$y = mx + 4 \sqrt{1 + m^2} \tag{2}$$

2. Substitute the Point $(0, 5)$: The point $(0, 5)$ lies on the tangent, so substituting $x = 0$ and $y = 5$ into equation (2):

$$5 = 4 \sqrt{1 + m^2}$$

3. Solve for $m$:

$$\frac{5}{4} = \sqrt{1 + m^2}$$ $$\left( \frac{5}{4} \right)^2 = 1 + m^2$$ $$\frac{25}{16} = 1 + m^2$$ $$m^2 = \frac{9}{16}$$ $$m = \pm \frac{3}{4} \tag{4}$$

Thus, the slopes of the tangents are $m = \pm \frac{3}{4}$.

4. Equation of the Tangents: Substituting $m = \pm \frac{3}{4}$ into the equation for the tangent:

$$y = \pm \frac{3}{4}x + 4 \sqrt{1 + \frac{9}{16}}$$

Simplify:

$$y = \pm \frac{3}{4}x + 5$$

Multiply through by 4:

$$4y = \pm 3x + 20$$

Thus, the equations of the tangents are:

$$4y - 3x = 20 \quad \text{and} \quad 3x - 4y + 20 = 0$$

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ii. Tangents from $(-1, 2)$ to the Circle $x^2 + y^2 + 4x + 2y = 0$

1. Rewrite the Circle Equation: The equation of the circle is $x^2 + y^2 + 4x + 2y = 0$. Complete the square to put it into standard form:

$$(x+2)^2 + (y+1)^2 = 5$$

So, the center of the circle is $(-2, -1)$ and the radius is $\sqrt{5}$.

2. Use the Pythagorean Theorem: The point $(-1, 2)$ is outside the circle, so the distance from this point to the center $(-2, -1)$ is the length of the tangent. Using the distance formula:

$$\text{Distance} = \sqrt{(-1 + 2)^2 + (2 + 1)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$$

3. Equation of the Tangents: Use the equation of the tangent from an external point $(x_1, y_1)$ to a circle $(x - h)^2 + (y - k)^2 = r^2$, which is:

$$(x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2$$

Substituting the values:

$$(-1 - (-2))(x + 2) + (2 - (-1))(y + 1) = 5$$

Simplifying:

$$(x + 2) + 3(y + 1) = 5$$

Thus, the equation of the tangents is:

$$x + 3y + 5 = 0$$

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iii. Tangents from $(-7, -2)$ to the Circle $(x + 1)^2 + (y - 2)^2 = 26$

1. Equation of the Circle: The center of the circle is $(-1, 2)$ and the radius is $\sqrt{26}$.

2. Equation of the Tangents: The general equation of the tangent from an external point $(x_1, y_1)$ to a circle $(x - h)^2 + (y - k)^2 = r^2$ is:

$$(x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2$$

Substituting the values:

$$(-7 + 1)(x + 1) + (-2 - 2)(y - 2) = 26$$

Simplify:

$$-6(x + 1) - 4(y - 2) = 26$$

Expanding and simplifying:

$$-6x - 6 - 4y + 8 = 26$$

Thus, the equation of the tangents is:

$$6x + 4y = 12$$

Simplifying further:

$$3x + 2y = 6$$

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Key Formulas or Methods Used

• General Equation of a Tangent to a Circle: $y = mx + c$
• Condition for Tangency: The perpendicular distance from the center of the circle to the tangent must equal the radius.
• Pythagorean Theorem: Used to calculate the distance between a point and the center of the circle.

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Summary of Steps

1. For Tangents from $(0, 5)$ to $x^2 + y^2 = 16$:

   • Solve for the slope $m$ using the condition that the point $(0, 5)$ lies on the tangent.
   • Find the equations of the tangents as $4y - 3x = 20$ and $3x - 4y + 20 = 0$.

2. For Tangents from $(-1, 2)$ to $x^2 + y^2 + 4x + 2y = 0$:

   • Rewrite the equation of the circle in standard form and find the radius.
   • Use the formula for the tangent from an external point to find the equation $x + 3y + 5 = 0$.

3. For Tangents from $(-7, -2)$ to $(x + 1)^2 + (y - 2)^2 = 26$:

   • Use the general tangent formula and simplify to get the equation $3x + 2y = 6$.