Circles Centred at the Origin

A circle is every point that’s the same distance from its centre. Put the centre at the origin, use the length formula, and you get a short, neat equation: $x^2 + y^2 = r^2$. With it you can sketch a circle in seconds, tell whether a point is inside or outside, and solve problems about broadcast ranges, diameters and chords.

Key ideas

Developing the equation

Let $P(x, y)$ be any point on a circle with centre $O(0, 0)$ and radius $r$. The distance from $O$ to $P$ is always $r$, so by the length formula:

$$\begin{aligned} \sqrt{(x - 0)^2 + (y - 0)^2} &= r \\ \sqrt{x^2 + y^2} &= r \\ x^2 + y^2 &= r^2 && \text{square both sides} \end{aligned}$$

So the equation of a circle with centre $(0, 0)$ and radius $r$ is

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

A point is on the circle exactly when its coordinates make this equation true.

Reading the radius from the equation

The number on the right is $r^2$, not $r$. Take the square root to get the radius:

• $x^2 + y^2 = 49$ has $r = \sqrt{49} = 7$.
• $x^2 + y^2 = 20$ has $r = \sqrt{20} \approx 4.47$.

Writing the equation

• Given the radius: square it. A radius of $6$ gives $x^2 + y^2 = 36$.
• Given a point on the circle: the distance from the origin to that point is the radius, so substitute the point to get $r^2$. Through $(3, -4)$: $r^2 = 3^2 + (-4)^2 = 25$, so $x^2 + y^2 = 25$.

Sketching

Plot the centre $(0, 0)$, then the four points one radius away along the axes: $(r, 0)$, $(-r, 0)$, $(0, r)$ and $(0, -r)$. Join them with a smooth round curve. A few extra points (like $(3, 4)$ on $x^2 + y^2 = 25$) help you keep it round. You can check a sketch with graphing technology such as Desmos.

On, inside, or outside?

For a point $(a, b)$ and the circle $x^2 + y^2 = r^2$, work out $a^2 + b^2$ (the point’s squared distance from the centre) and compare it with $r^2$:

If $a^2 + b^2$ is …	the point is …
equal to $r^2$	on the circle
less than $r^2$	inside the circle
greater than $r^2$	outside the circle

Diameters and chords

A chord is a segment joining two points on a circle. A diameter is a chord through the centre. For a circle centred at the origin, the endpoints of a diameter are opposite each other through the origin: if one endpoint is $(a, b)$, the other is $(-a, -b)$, and their midpoint is $(0, 0)$.

A key property: the right bisector of any chord (the line perpendicular to the chord through its midpoint) passes through the centre of the circle. Example 4 checks this with coordinates. You’ll write right bisector equations in more detail on right bisectors and distance.

Worked examples

Example 1: Radius from the equation

State the radius of each circle, and the points where it crosses the axes.

• (a) $x^2 + y^2 = 36$
• (b) $x^2 + y^2 = 18$

Solution.

(a) $r^2 = 36$, so $r = \sqrt{36} = 6$. The circle crosses the axes at $(6, 0)$, $(-6, 0)$, $(0, 6)$ and $(0, -6)$.

(b) $r^2 = 18$, so $r = \sqrt{18} \approx 4.24$ (exactly, $\sqrt{18} = 3\sqrt{2}$). The circle crosses the axes at about $(4.24, 0)$, $(-4.24, 0)$, $(0, 4.24)$ and $(0, -4.24)$.

Example 2: Writing the equation

Write the equation of the circle with centre $(0, 0)$ that

• (a) has radius $9$
• (b) passes through the point $(-2, 5)$

Solution.

(a) $r^2 = 9^2 = 81$, so the equation is $x^2 + y^2 = 81$.

(b) The radius is the distance from $(0, 0)$ to $(-2, 5)$, so

$$r^2 = (-2)^2 + 5^2 = 4 + 25 = 29$$

The equation is $x^2 + y^2 = 29$, and the radius is $\sqrt{29} \approx 5.39$.

You don’t need to find $r$ itself to write the equation: $r^2$ is what goes in.

Example 3: Testing points and finding a diameter

For the circle $x^2 + y^2 = 25$:

• (a) Decide whether each point is on, inside or outside the circle: $(3, -4)$, $(2, 4)$, $(-4, 4)$.
• (b) One endpoint of a diameter is $(3, -4)$. Find the other endpoint.

Solution.

(a) Compare each $x^2 + y^2$ with $r^2 = 25$:

$$\begin{aligned} (3, -4)&: \quad 9 + 16 = 25 && \text{equal, so on the circle} \\ (2, 4)&: \quad 4 + 16 = 20 && \text{less than } 25 \text{, so inside} \\ (-4, 4)&: \quad 16 + 16 = 32 && \text{greater than } 25 \text{, so outside} \end{aligned}$$

(b) The centre $(0, 0)$ is the midpoint of the diameter, so the other endpoint is opposite $(3, -4)$ through the origin: $(-3, 4)$.

Check: the midpoint of $(3, -4)$ and $(-3, 4)$ is $(0, 0)$, and $(-3)^2 + 4^2 = 25$, so $(-3, 4)$ is on the circle. ✓

Example 4: A chord and its right bisector

The points $A(-5, 0)$ and $B(3, 4)$ are on the circle $x^2 + y^2 = 25$. Find the equation of the right bisector of chord $AB$, and show that it passes through the centre.

Solution. First check that both points are on the circle: $(-5)^2 + 0^2 = 25$ ✓ and $3^2 + 4^2 = 25$ ✓.

Midpoint of $AB$:

$$M = \left( \frac{-5 + 3}{2},\ \frac{0 + 4}{2} \right) = (-1, 2)$$

Slope of $AB$:

$$m_{AB} = \frac{4 - 0}{3 - (-5)} = \frac{4}{8} = \frac{1}{2}$$

The right bisector is perpendicular to $AB$, so its slope is the negative reciprocal, $-2$.

Equation: substitute $M(-1, 2)$ into $y = -2x + b$:

$$2 = -2(-1) + b \quad\Rightarrow\quad 2 = 2 + b \quad\Rightarrow\quad b = 0$$

The right bisector is $y = -2x$. Its $y$-intercept is $0$, so it passes through $(0, 0)$, the centre of the circle. ✓

The right bisector of chord $AB$, $y = -2x$, passes through the centre of the circle.

Common mistakes

Using $r^2$ as the radius. In $x^2 + y^2 = 36$, the radius is $6$, not $36$. The right side is the radius squared. Take the square root.

Forgetting to square the radius. A circle with radius $9$ is $x^2 + y^2 = 81$, not $x^2 + y^2 = 9$. Substitute a point you know, like $(9, 0)$, to check: $81 + 0 = 81$. ✓

Squaring a negative coordinate wrongly. $(-4)^2 = 16$, not $-16$. Squared distances are never negative. Use brackets when you substitute.

Comparing with $r$ instead of $r^2$. For $x^2 + y^2 = 25$ and the point $(2, 4)$, compare $20$ with $25$, not with $5$. Either compare $a^2 + b^2$ with $r^2$, or compare the actual distance $\sqrt{20} \approx 4.47$ with $r = 5$. Don’t mix them.

Using the chord’s slope for the right bisector. The right bisector is perpendicular to the chord, so use the negative reciprocal of the chord’s slope, and make sure the line goes through the chord’s midpoint, not one of its endpoints.

Practice

1. (Warm-up) State the radius of each circle. Give an exact value and, where needed, a decimal to two decimal places.

• (a) $x^2 + y^2 = 64$
• (b) $x^2 + y^2 = 1.44$
• (c) $x^2 + y^2 = 12$

Solution

(a) $r = \sqrt{64} = 8$

(b) $r = \sqrt{1.44} = 1.2$

(c) $r = \sqrt{12} \approx 3.46$ (exactly, $\sqrt{12} = 2\sqrt{3}$)

2. (Warm-up) Write the equation of the circle with centre $(0, 0)$ and the given radius.

• (a) $11$
• (b) $\sqrt{7}$
• (c) $2.5$

Solution

(a) $x^2 + y^2 = 121$

(b) $\left(\sqrt{7}\right)^2 = 7$, so $x^2 + y^2 = 7$

(c) $2.5^2 = 6.25$, so $x^2 + y^2 = 6.25$

3. (Core) A circle with centre $(0, 0)$ passes through $(6, -8)$. Write its equation, state its radius, and list the four points where it crosses the axes (to help you sketch it).

Solution

$$r^2 = 6^2 + (-8)^2 = 36 + 64 = 100$$

The equation is $x^2 + y^2 = 100$, and $r = 10$. It crosses the axes at $(10, 0)$, $(-10, 0)$, $(0, 10)$ and $(0, -10)$.

4. (Core) For the circle $x^2 + y^2 = 50$, decide whether each point is on, inside or outside the circle.

• (a) $(5, 5)$
• (b) $(-7, 1)$
• (c) $(6, -4)$
• (d) $(0, 7)$

Solution

Compare $x^2 + y^2$ with $50$.

(a) $25 + 25 = 50$: on the circle.

(b) $49 + 1 = 50$: on the circle.

(c) $36 + 16 = 52$: greater than $50$, so outside.

(d) $0 + 49 = 49$: less than $50$, so inside (just barely).

5. (Core) One endpoint of a diameter of a circle centred at the origin is $(-7, 24)$. Find the other endpoint and the equation of the circle.

Solution

The other endpoint is opposite through the origin: $(7, -24)$.

$$r^2 = (-7)^2 + 24^2 = 49 + 576 = 625$$

The equation is $x^2 + y^2 = 625$ (radius $25$).

6. (Core) The point $(k, -3)$ is on the circle $x^2 + y^2 = 34$. Find all possible values of $k$.

Solution

$$\begin{aligned} k^2 + (-3)^2 &= 34 \\ k^2 + 9 &= 34 \\ k^2 &= 25 \\ k &= 5 \quad \text{or} \quad k = -5 \end{aligned}$$

There are two points: $(5, -3)$ and $(-5, -3)$.

7. (Core) A radio station’s tower is at the origin of a map grid measured in kilometres. Its signal reaches $40$ km in every direction.

• (a) Write an equation for the edge of the broadcast area.
• (b) Can a town at $(25, 30)$ receive the signal? What about a town at $(-32, 26)$?

Solution

(a) $r = 40$, so the edge is $x^2 + y^2 = 1600$.

(b) Town at $(25, 30)$: $25^2 + 30^2 = 625 + 900 = 1525$, which is less than $1600$. It’s inside the circle, so it gets the signal (it’s about $\sqrt{1525} \approx 39.1$ km away).

Town at $(-32, 26)$: $(-32)^2 + 26^2 = 1024 + 676 = 1700$, which is more than $1600$. It’s outside, so it doesn’t get the signal.

8. (Challenge) The points $P(-1, 7)$ and $Q(5, 5)$ are on the circle $x^2 + y^2 = 50$. Show that the right bisector of chord $PQ$ passes through the centre of the circle.

Solution

Both points are on the circle: $1 + 49 = 50$ ✓ and $25 + 25 = 50$ ✓.

Midpoint of $PQ$: $\left(\dfrac{-1 + 5}{2}, \dfrac{7 + 5}{2}\right) = (2, 6)$.

Slope of $PQ$: $\dfrac{5 - 7}{5 - (-1)} = \dfrac{-2}{6} = -\dfrac{1}{3}$, so the perpendicular slope is $3$.

Substitute $(2, 6)$ into $y = 3x + b$: $6 = 6 + b$, so $b = 0$.

The right bisector is $y = 3x$. Its $y$-intercept is $0$, so it passes through the centre $(0, 0)$. ✓

9. (Challenge) $A(-5, 0)$ and $B(5, 0)$ are the endpoints of a diameter of the circle $x^2 + y^2 = 25$, and $C(3, 4)$ is on the circle. Show that triangle $ABC$ has a right angle at $C$.

Solution

$C$ is on the circle because $9 + 16 = 25$.

$$m_{AC} = \frac{4 - 0}{3 - (-5)} = \frac{4}{8} = \frac{1}{2} \qquad m_{BC} = \frac{4 - 0}{3 - 5} = \frac{4}{-2} = -2$$

The slopes are negative reciprocals (their product is $\dfrac{1}{2} \times (-2) = -1$), so $AC \perp BC$. Triangle $ABC$ has a right angle at $C$.

(This is true for any point $C$ on the circle: an angle drawn from the ends of a diameter to the circle is always a right angle.)