6.1 Q-1

Question Statement

In this problem, you are tasked with finding the equation of a circle given certain details about its center and radius. The problem is broken into three parts:

Part (a): A circle with center at $(5, -2)$ and radius $4$ .
Part (b): A circle with center at $\left(\sqrt{2}, -3\sqrt{3}\right)$ and radius $2\sqrt{2}$ .
Part (c): A circle with the endpoints of its diameter at $(-3, 2)$ and $(5, -6)$ .

Background and Explanation

To solve this problem, we need to recall the general equation of a circle:

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

Where:

$(h, k)$ represents the center of the circle.
$r$ is the radius.

In some cases, you may need to expand this equation into its general form:

x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0

For Part (c), where you’re given the endpoints of the diameter, you’ll first find the midpoint to get the center of the circle. The radius will be half the distance between the two endpoints.

Solution

Part (a)

Given: Center at $(5, -2)$ and radius $4$ .

Step 1: Use the standard circle equation:

(x - 5)^2 + (y + 2)^2 = 4^2

Step 2: Expand both squared terms:

(x - 5)^2 = x^2 - 10x + 25 \quad \text{and} \quad (y + 2)^2 = y^2 + 4y + 4

Step 3: Substitute the expansions into the equation:

x^2 - 10x + 25 + y^2 + 4y + 4 = 16

Step 4: Combine like terms and move the constant to the left side:

x^2 + y^2 - 10x + 4y + 25 + 4 - 16 = 0

Step 5: Simplify the equation:

x^2 + y^2 - 10x + 4y + 13 = 0

Thus, the equation of the circle is:

x^2 + y^2 - 10x + 4y + 13 = 0

Part (b)

Given: Center at $\left( \sqrt{2}, -3\sqrt{3} \right)$ and radius $2\sqrt{2}$ .

Step 1: Use the standard equation:

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

Step 2: Expand the squared terms:

(x - \sqrt{2})^2 = x^2 - 2\sqrt{2}x + 2 \quad \text{and} \quad (y + 3\sqrt{3})^2 = y^2 + 6\sqrt{3}y + 27

Step 3: Substitute these into the equation:

x^2 - 2\sqrt{2}x + 2 + y^2 + 6\sqrt{3}y + 27 = 8

Step 4: Combine like terms and move the constant to the left side:

x^2 + y^2 - 2\sqrt{2}x + 6\sqrt{3}y + 2 + 27 - 8 = 0

Step 5: Simplify the equation:

x^2 + y^2 - 2\sqrt{2}x + 6\sqrt{3}y + 21 = 0

Thus, the equation of the circle is:

x^2 + y^2 - 2\sqrt{2}x + 6\sqrt{3}y + 21 = 0

Part (c)

Given: Ends of the diameter at $(-3, 2)$ and $(5, -6)$ .

Step 1: Find the center by calculating the midpoint of the diameter:

\text{Midpoint} = \left( \frac{-3 + 5}{2}, \frac{2 + (-6)}{2} \right) = \left( \frac{2}{2}, \frac{-4}{2} \right) = (1, -2)

Step 2: Calculate the radius using the distance formula (distance from the center to one endpoint):

r = \frac{1}{2} \sqrt{(5 - (-3))^2 + (-6 - 2)^2} = \frac{1}{2} \sqrt{8^2 + (-8)^2} = \frac{1}{2} \sqrt{64 + 64} = \frac{1}{2} \sqrt{128} = 4\sqrt{2}

Step 3: Use the standard equation of the circle:

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

Step 4: Expand the squared terms:

(x - 1)^2 = x^2 - 2x + 1 \quad \text{and} \quad (y + 2)^2 = y^2 + 4y + 4

Step 5: Substitute these into the equation:

x^2 - 2x + 1 + y^2 + 4y + 4 = 32

Step 6: Combine like terms and move the constant to the left side:

x^2 + y^2 - 2x + 4y + 1 + 4 - 32 = 0

Step 7: Simplify the equation:

x^2 + y^2 - 2x + 4y - 27 = 0

Thus, the equation of the circle is:

x^2 + y^2 - 2x + 4y - 27 = 0

Key Formulas or Methods Used

Equation of a Circle:
$(x - h)^2 + (y - k)^2 = r^2$
The standard form of a circle, where $(h, k)$ is the center and $r$ is the radius.
Midpoint Formula (to find the center from the endpoints of a diameter):
$\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
Distance Formula (to find the radius from the center and an endpoint):
$r = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Summary of Steps

Part (a): Write the equation using the standard form, expand, and simplify.
Part (b): Use the standard form, expand the terms, and simplify.
Part (c): Calculate the midpoint (center) and radius, then use the standard equation to form and simplify the equation.