Semicircles and Circles from a Graph

Start from $x^2 + y^2 = r^2$ and make $y$ the subject: $y^2 = r^2 - x^2$, so $y = \pm\sqrt{r^2 - x^2}$.

The plus-or-minus is the whole story. Choosing the positive root keeps only points with $y \geq 0$, which is the upper semicircle. Choosing the negative root keeps $y \leq 0$, the lower one.

For $r = 5$: $y = \sqrt{25 - x^2}$ is the upper half and $y = -\sqrt{25 - x^2}$ the lower. Together they reassemble the full circle.

Solving $x^2+y^2=r^2$ for $y$ gives $y = \pm\sqrt{r^2-x^2}$; the positive root is the upper semicircle and the negative root the lower. The square root symbol alone means the positive root, so the plus-or-minus must be written deliberately.

Pause, copy the rearrangement to $y = \pm\sqrt{r^2-x^2}$, which sign gives which half, and the note that $\sqrt{\ }$ alone is positive, into your book.

We just saw how the two roots split the circle. That raises a question: what happens to the domain and range once you keep only half? This card answers it → the domain is unchanged and the range is halved.

The upper semicircle $y = \sqrt{r^2 - x^2}$ still stretches from $x = -r$ to $x = r$, so its domain is $[-r, r]$, exactly the same as the full circle.

Its range, though, is only $[0, r]$, because the positive root can never be negative. The lower semicircle has the same domain with range $[-r, 0]$.

The right and left semicircles, from solving for $x$, swap these around: domain $[0, r]$ or $[-r, 0]$, with range $[-r, r]$ in both cases.

An upper or lower semicircle keeps the full domain $[-r,r]$ but halves the range to $[0,r]$ or $[-r,0]$. Left and right semicircles swap those. Each semicircle passes the vertical line test, so each is a function while the whole circle is not.

Pause, copy the domain and range of all four semicircles, and the reason each one is a function, into your book.

We just saw how to go from an equation to a half-circle. That raises a question: can you run the whole thing backwards, starting from a picture? This card answers it → yes, and for a circle centred at the origin you need only one number.

A circle centred at the origin is determined entirely by its radius, so read the radius from any axis crossing. If the curve passes through $(6, 0)$, the radius is 6.

Square it for the right-hand side: $x^2 + y^2 = 36$. That is the whole answer.

If the graph is only a half, decide which one and add the matching root. An upper half of radius 6 is $y = \sqrt{36 - x^2}$; a left half is $x = -\sqrt{36 - y^2}$.

A circle centred at the origin needs only its radius. Read it from an axis crossing, or compute $r^2 = x^2 + y^2$ from any point on the curve, then square for the right-hand side. For a half-graph, choose the matching root.

Pause, copy the read-the-radius method, the worked $(3,4)$ giving $x^2+y^2=25$, and how to add the correct root for a semicircle, into your book.

1. Solve $x^2 + y^2 = 9$ for $y$.
2. Which semicircle is $y = -\sqrt{16 - x^2}$?
3. State the range of $y = \sqrt{4 - x^2}$.
4. A circle centred at the origin passes through $(0, 7)$. Find its equation.