6.1 Q-9

Question Statement

Find the equation(s) of the circle with a radius of 2 that is tangent to the line $x - y - 4 = 0$ at the point $A(1, -3)$.

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Background and Explanation

To solve this problem, we need to find the equation of a circle with the following properties:

• The radius is 2.
• The circle is tangent to the given line $x - y - 4 = 0$ at the point $A(1, -3)$.

Key Concepts:

• Equation of a Circle: The general form of a circle is:

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

where $(h, k)$ is the center and $r$ is the radius.

• Tangent Line: For a circle to be tangent to a line, the radius at the point of tangency must be perpendicular to the line.

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Solution

Step 1: Set up the equation of the circle

Let the center of the circle be $C(h, k)$, and the radius be $r = 2$. The circle passes through the point $A(1, -3)$, so the equation of the circle is:

$$(h - 1)^2 + (k + 3)^2 = 2^2$$

Expanding:

$$h^2 - 2h + 1 + k^2 + 6k + 9 = 4$$

Simplifying:

$$h^2 + k^2 - 2h + 6k + 6 = 0 \tag{1}$$

Step 2: Use the condition of tangency

The line $x - y - 4 = 0$ has a slope of 1. The line through $A(1, -3)$ and $C(h, k)$ must also be perpendicular to this line. The slope of the line through $A(1, -3)$ and $C(h, k)$ is:

$$\frac{k + 3}{h - 1}$$

Since the two lines are perpendicular, their slopes must satisfy:

$$1 \times \frac{k + 3}{h - 1} = -1$$

This simplifies to:

$$k + 3 = -h + 1 \quad \Rightarrow \quad k = -h - 2 \tag{2}$$

Step 3: Substitute $k = -h - 2$ into equation (1)

Substitute $k = -h - 2$ from equation (2) into the equation for the circle (1):

$$h^2 + (-h - 2)^2 - 2h + 6(-h - 2) + 6 = 0$$

Expanding:

$$h^2 + (h^2 + 4h + 4) - 2h - 6h - 12 + 6 = 0$$

Simplifying:

$$2h^2 - 4h - 2 = 0$$

Divide the entire equation by 2:

$$h^2 - 2h - 1 = 0$$

Step 4: Solve for $h$

Solve the quadratic equation $h^2 - 2h - 1 = 0$ using the quadratic formula:

$$h = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2}$$

Simplifying:

$$h = 1 \pm \sqrt{2}$$

Thus, there are two possible values for $h$:

• $h = 1 + \sqrt{2}$
• $h = 1 - \sqrt{2}$

Step 5: Solve for $k$

Using equation (2), $k = -h - 2$, we can find the corresponding values for $k$:

• When $h = 1 + \sqrt{2}$, then:

$$k = -1 - \sqrt{2} - 2 = -3 - \sqrt{2}$$

• When $h = 1 - \sqrt{2}$, then:

$$k = -1 + \sqrt{2} - 2 = -3 + \sqrt{2}$$

Step 6: Write the equations of the circles

For each pair of values for $(h, k)$, we can write the equation of the circle:

1. For $h = 1 + \sqrt{2}, k = -3 - \sqrt{2}$, the equation of the circle is:

$$(x - (1 + \sqrt{2}))^2 + (y + (3 + \sqrt{2}))^2 = 4$$

2. For $h = 1 - \sqrt{2}, k = -3 + \sqrt{2}$, the equation of the circle is:

$$(x - (1 - \sqrt{2}))^2 + (y + (3 - \sqrt{2}))^2 = 4$$

Both circles have a radius of 2 and are tangent to the line $x - y - 4 = 0$ at the point $A(1, -3)$.

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Key Formulas or Methods Used

• Equation of a Circle:
  The general form of the equation of a circle is:

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

• Slope of a Line:
  The slope of the line through two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$$\text{Slope} = \frac{y_2 - y_1}{x_2 - x_1}$$

• Condition for Perpendicular Lines:
  If two lines are perpendicular, the product of their slopes is $-1$.

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Summary of Steps

1. Step 1: Set up the equation of the circle using the given point $A(1, -3)$ and radius $r = 2$.
2. Step 2: Use the condition of tangency to relate the slope of the tangent line and the slope of the line from $A(1, -3)$ to the center.
3. Step 3: Substitute $k = -h - 2$ into the equation of the circle and simplify.
4. Step 4: Solve the quadratic equation for $h$ and find the corresponding values for $k$.
5. Step 5: Write the equations of the two circles based on the values of $h$ and $k$.