Circles and Semicircles Revision Notes for HSC SSCE Mathematics Advanced

Circles and Semicircles

A circle is a set of points that are all the same distance from a fixed centre point. In mathematics, we can describe circles using equations, and understanding these equations helps us graph circles and work with their properties.

Deriving the circle equation

The equation of a circle is found by applying Pythagoras' theorem to describe all points at equal distance from a centre point.

Consider any point $(x, y)$ on a circle with its centre at the origin $(0, 0)$ . This point is at a specific distance $r$ (the radius) from the centre.

A right-angled triangle is formed with:

Vertices at $(0, 0)$ , $(x, 0)$ , and $(x, y)$
The hypotenuse is the radius $r$
The horizontal leg has length $x$
The vertical leg has length $y$

Using Pythagoras' theorem, we can write:

$x^2 + y^2 = r^2$

Where:

$x$ is the x-coordinate of a point on the circle
$y$ is the y-coordinate of a point on the circle
$r$ is the radius (distance from the origin to any point on the circumference)

chatImportant

This equation $x^2 + y^2 = r^2$ ensures that every point satisfying it is exactly distance $r$ from the origin. The value of $r$ determines the circle's size - a larger radius produces a larger circle.

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Worked Example: Deriving a circle equation

Question: Derive the equation of a circle with a distance of 7 from its fixed point at the origin.

Solution:

Start with the circle equation:

$x^2 + y^2 = r^2$

Substitute $r = 7$ :

$x^2 + y^2 = 7^2$

Evaluate:

$x^2 + y^2 = 49$

The equation is x² + y² = 49.

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Worked Example: Verifying a point on a circle

Question: Verify that the point $(3, 4)$ lies on a circle with a distance of 5 from its fixed point at the origin.

Solution:

Write the equation:

$x^2 + y^2 = r^2$

Substitute $x = 3$ , $y = 4$ , and $r = 5$ :

$3^2 + 4^2 = 5^2$

Evaluate each term:

$9 + 16 = 25$

Evaluate:

$25 = 25$

Since both sides are equal, the point $(3, 4)$ satisfies $x^2 + y^2 = 25$ , confirming it lies on the circle.

Graphing circles

To graph a circle with equation $x^2 + y^2 = r^2$ , we plot the centre at the origin $(0, 0)$ and mark points at distance $r$ along the axes.

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The key features of a circle centred at the origin are:

Centre at $(0, 0)$
Radius $r$
x-intercepts at $(\pm r, 0)$
y-intercepts at $(0, \pm r)$

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Worked Example: Graphing a circle

Question: Graph the circle $x^2 + y^2 = 16$ .

Solution:

First, identify the radius. The equation $x^2 + y^2 = 16$ has the form $x^2 + y^2 = r^2$ , so:

$r^2 = 16$

$r = \sqrt{16} = 4$

The circle has:

Centre at $(0, 0)$
Radius $r = 4$
Key points at $(4, 0)$ , $(-4, 0)$ , $(0, 4)$ , and $(0, -4)$

Check: We can verify the intercepts by substituting. When $x = 0$ : $y^2 = 16$ , so $y = \pm 4$ . When $y = 0$ : $x^2 = 16$ , so $x = \pm 4$ . This confirms our intercepts.

Finding the equation from a graph

When we have a graph of a circle, we can determine its equation by finding the radius. The circle is centred at the origin $(0, 0)$ , and we find $r$ by measuring the distance from the origin to any point on the circumference.

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The intercepts are often the easiest points to use since they lie on the axes, making calculations simpler.

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Worked Example: Finding equation from a point

Question: A circle has its fixed point at $(0, 0)$ and passes through $(0, 6)$ . Find its equation.

Solution:

We need to find the radius $r$ . Substitute the point $(0, 6)$ into $x^2 + y^2 = r^2$ :

$x^2 + y^2 = r^2$

$0^2 + 6^2 = r^2$

$36 = r^2$

$r = 6$

(taking the positive square root)

The equation is x² + y² = 36.

Check: The point $(6, 0)$ should also satisfy this equation: $6^2 + 0^2 = 36$ , which is true.

Semicircles

A semicircle is half of a circle. When we solve the circle equation $x^2 + y^2 = r^2$ for one variable, we create equations for semicircles.

Upper and lower semicircles

Solving the circle equation for $y$ gives:

$y = \pm\sqrt{r^2 - x^2}$

This produces two functions:

Upper semicircle: $y = \sqrt{r^2 - x^2}$ (positive square root, above the x-axis)
Lower semicircle: $y = -\sqrt{r^2 - x^2}$ (negative square root, below the x-axis)

For these semicircles:

$y$ is the y-coordinate of a point on the semicircle
$x$ is the x-coordinate of a point on the semicircle
$r$ is the radius
Domain: $-r \leq x \leq r$ or $[-r, r]$

Right and left semicircles

Solving the circle equation for $x$ gives:

$x = \pm\sqrt{r^2 - y^2}$

This produces two functions:

Right semicircle: $x = \sqrt{r^2 - y^2}$ (positive square root, right of the y-axis)
Left semicircle: $x = -\sqrt{r^2 - y^2}$ (negative square root, left of the y-axis)

For these semicircles:

$x$ is the x-coordinate of a point on the semicircle
$y$ is the y-coordinate of a point on the semicircle
$r$ is the radius
Range: $-r \leq y \leq r$ or $[-r, r]$

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Key pattern to remember:

Positive square root = Upper or Right semicircle
Negative square root = Lower or Left semicircle

The sign of the square root determines which half of the circle you get!

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Understanding Domain and Range:

For upper/lower semicircles (solved for $y$ ), the domain tells you how far left and right the semicircle extends along the x-axis.

For right/left semicircles (solved for $x$ ), the range tells you how far up and down the semicircle extends along the y-axis.

Example: The equation $y = \sqrt{r^2 - x^2}$ with radius $r = 5$ has domain $[-5, 5]$ and forms the upper semicircle, while $x = -\sqrt{r^2 - y^2}$ with $r = 4$ has range $[-4, 4]$ and forms the left semicircle.

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Worked Example: Graphing an upper semicircle

Question: Graph the semicircle $y = \sqrt{25 - x^2}$ .

Solution:

First, identify the form and radius. The equation comes from $x^2 + y^2 = 25$ , so:

$r^2 = 25$

$r = 5$

The positive square root indicates this is the upper semicircle.

Find the domain. For the square root to be defined:

$25 - x^2 \geq 0$

$x^2 \leq 25$

$-5 \leq x \leq 5$

Domain: $[-5, 5]$

Key points to plot:

$(0, 5)$ - the highest point
$(5, 0)$ - right endpoint
$(-5, 0)$ - left endpoint
$(-3, 4)$ and $(3, 4)$ - additional points for accuracy

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Worked Example: Graphing a left semicircle

Question: Graph the semicircle $x = -\sqrt{16 - y^2}$ .

Solution:

Identify the form and radius. The equation comes from $x^2 + y^2 = 16$ , so $r = 4$ .

The negative square root indicates this is the left semicircle.

Find the range. For the square root to be defined:

$16 - y^2 \geq 0$

$y^2 \leq 16$

$-4 \leq y \leq 4$

Range: $[-4, 4]$

Key points to plot:

$(-4, 0)$ - the leftmost point
$(0, 4)$ - top endpoint
$(0, -4)$ - bottom endpoint
$(-3, 2.645)$ and $(-3, -2.645)$ - additional points

Check: The point $(-3, 2.645)$ approximately satisfies the original circle equation: $(-3)^2 + (2.645)^2 \approx 9 + 7 = 16$ , confirming it lies on the semicircle.

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Key Points to Remember:

The equation of a circle centred at the origin is x² + y² = r², derived from Pythagoras' theorem
To graph a circle, plot the centre at $(0, 0)$ and mark points at distance $r$ along the x- and y-axes
To find a circle's equation from its graph, identify the radius from a point on the circumference
Semicircles are created by solving for one variable: $y = \pm\sqrt{r^2 - x^2}$ gives upper/lower semicircles; $x = \pm\sqrt{r^2 - y^2}$ gives right/left semicircles
The positive square root gives the upper or right semicircle, while the negative square root gives the lower or left semicircle

Circles and Semicircles (HSC SSCE Mathematics Advanced): Revision Notes