Circles-Their Equations and etc.

Parent Function of a Circle- x²+y²=1

1 being the square of the radius

X and Y being the coordinates of the center

Parent Function of a Circle: $x^2 + y^2 = 1$

1. Understanding the Circle Equation

The parent function of a circle is centered at the origin $(0, 0)$ with a radius of $1$ unit (also known as the unit circle).

• Standard Form Equation: $(x - h)^2 + (y - k)^2 = r^2$

  • $(h, k)$: Coordinates of the center of the circle.

  • $r$: Radius of the circle ($r > 0$).

  • $r^2$: The square of the radius.

• Parent Function Parameters:

  • Center: $(h, k) = (0, 0)$

  • Radius: $r = 1$

  • Equation: $(x - 0)^2 + (y - 0)^2 = 1^2 → x^2 + y^2 = 1$

  • Domain: $[-1, 1]$

  • Range: $[-1, 1]$

2. Derivation from the Distance Formula

A circle is defined as the set of all points $(x, y)$ in a plane that are equidistant from a fixed center point $(h, k)$.

Using the distance formula between $(x, y)$ and $(h, k)$:
$\text{Distance} = \sqrt{(x - h)^2 + (y - k)^2}$

Setting the distance equal to the radius $r$:
$r = \sqrt{(x - h)^2 + (y - k)^2}$

Squaring both sides yields standard form:
$(x - h)^2 + (y - k)^2 = r^2$

3. Worked Examples

Example 1: Finding Center and Radius

Problem: Identify the center and radius of the circle given by $(x - 3)^2 + (y + 4)^2 = 25$.

Solution:

1. Compare the equation to standard form $(x - h)^2 + (y - k)^2 = r^2$:

   • $(x - 3)^2 \implies h = 3$

   • $(y - (-4))^2 \implies k = -4$

   • $r^2 = 25 \implies r = \sqrt{25} = 5$

2. Result:

   • Center: $(3, -4)$

   • Radius: $5$

Example 2: Writing the Equation

Problem: Write the equation of a circle centered at $(-2, 5)$ with a radius of $2\sqrt{3}$.

Solution:

1. Substitute $h = -2$, $k = 5$, and $r = 2\sqrt{3}$ into standard form:

   • $(x - (-2))^2 + (y - 5)^2 = (2\sqrt{3})^2$

2. Simplify:

   • $(x + 2)^2 + (y - 5)^2 = 12$

Example 3: Completing the Square

Problem: Convert $x^2 + y^2 - 6x + 8y + 9 = 0$ into standard form to find the center and radius.

Solution:

1. Group $x$ and $y$ terms, and move constant term:

   • $(x^2 - 6x) + (y^2 + 8y) = -9$

2. Complete the square for $x$ and $y$:

   • For $x$: \left(\frac{-6}{2}\right)^2 = 9

   • For $y$: \left(\frac{8}{2}\right)^2 = 16

   • Add $9$ and $16$ to both sides:

   • $(x^2 - 6x + 9) + (y^2 + 8y + 16) = -9 + 9 + 16$

3. Factor into standard form:

   • $(x - 3)^2 + (y + 4)^2 = 16$

4. Result:

   • Center: $(3, -4)$

   • Radius: $r = \sqrt{16} = 4$