Equation of a Circle

1 The Equation of a Circle Centred at the Origin

A circle centred at the origin $(0,0)$ with radius $r$ has equation: $$x^2 + y^2 = r^2$$ Every point $(x,y)$ that lies on the circle satisfies this equation — and only those points.

The Circle Equation

$$\boxed{x^2 + y^2 = r^2}$$

Centre $= (0,0)$, Radius $= r$ (always take the positive square root)

Where does $x^2+y^2=r^2$ come from?

For any point P$(x,y)$ on the circle, Pythagoras gives $x^2 + y^2 = r^2$.

Why does this work?
Drop a perpendicular from any point $P(x,y)$ on the circle to the x-axis. This creates a right-angled triangle with:

Horizontal leg of length $|x|$
Vertical leg of length $|y|$
Hypotenuse = radius $r$

By Pythagoras: $x^2 + y^2 = r^2$ ✓

💡 To find the radius from the equation, take the square root of the number on the right: $x^2+y^2=49 \Rightarrow r = \sqrt{49} = 7$

Worked Example 1 — Identifying Radius and Writing the Equation

①$x^2 + y^2 = 36$: Radius $= \sqrt{36} = \mathbf{6}$. Centre is the origin $(0,0)$.

②$x^2 + y^2 = 7$: Radius $= \sqrt{7} \approx 2.65$. (The radius doesn't have to be a whole number.)

③Circle with radius 9: Equation is $x^2 + y^2 = 81$ (since $9^2 = 81$).

④Circle with radius $\sqrt{5}$: Equation is $x^2 + y^2 = 5$.

Equation	Radius $r$	Check: does $(3,4)$ lie on it?
$x^2+y^2=25$	$\sqrt{25}=5$	$3^2+4^2=9+16=25$ ✓ Yes
$x^2+y^2=50$	$\sqrt{50}=5\sqrt{2}$	$9+16=25\neq50$ ✗ No
$x^2+y^2=169$	$\sqrt{169}=13$	$9+16=25\neq169$ ✗ No
$x^2+y^2=5$	$\sqrt{5}$	$9+16\neq5$ ✗ No

2 Points On, Inside, and Outside a Circle

To test whether a point $(a, b)$ is on, inside, or outside the circle $x^2+y^2=r^2$, substitute the coordinates and compare $a^2+b^2$ with $r^2$:

Result of $a^2 + b^2$	Position of point	Why?
$a^2+b^2 = r^2$	On the circle	Exactly radius $r$ from centre
$a^2+b^2 < r^2$	Inside the circle	Closer than $r$ to centre
$a^2+b^2 > r^2$	Outside the circle	Further than $r$ from centre

Testing three points against the circle $x^2+y^2=25$: green (inside), purple (on), red (outside).

Worked Example 2 — Classifying Points

For the circle $x^2+y^2=40$, state whether each point is on, inside, or outside:

①$(6,2)$: $\; 6^2+2^2 = 36+4 = 40 = 40$ → On the circle

②$(5,3)$: $\; 5^2+3^2 = 25+9 = 34 < 40$ → Inside the circle

③$(-7,1)$: $\; (-7)^2+1^2 = 49+1 = 50 > 40$ → Outside the circle

✓Always check your arithmetic — a single error will give the wrong classification.

Worked Example 3 — Finding a Missing Coordinate

The point $(k, 5)$ lies on the circle $x^2+y^2=61$. Find all possible values of $k$.

①Substitute into the equation: $k^2 + 5^2 = 61$

②$k^2 + 25 = 61$

③$k^2 = 36$

✓$k = \pm 6$, so the two points are $(6, 5)$ and $(-6, 5)$. Both lie on the circle — this makes sense by symmetry.

3 The Tangent to a Circle at a Given Point

A tangent to a circle is a straight line that touches the circle at exactly one point and does not cross it. The most important property is:

Key Property

The tangent at a point on a circle is perpendicular to the radius at that point.

If the radius has gradient $m$, the tangent has gradient $-\dfrac{1}{m}$ (since perpendicular lines satisfy $m_1 \times m_2 = -1$).

Tangent at $P(3,4)$ on $x^2+y^2=25$

For $P(3,4)$ on $x^2+y^2=25$:

Gradient of radius OP:
$m_{\text{radius}} = \dfrac{4-0}{3-0} = \dfrac{4}{3}$

Gradient of tangent:
$m_{\text{tangent}} = -\dfrac{1}{4/3} = -\dfrac{3}{4}$

(since radius $\perp$ tangent: $m_1 \times m_2 = -1$)

🔑 Perpendicular gradient: Flip the fraction and change the sign. If the radius gradient is $\dfrac{a}{b}$, the tangent gradient is $-\dfrac{b}{a}$.

4 Step-by-Step: Finding the Tangent Equation

Full Method — 4 Steps

To find the equation of the tangent to $x^2+y^2=r^2$ at point $P(x_1,y_1)$:

Verify that the point lies on the circle: check $x_1^2 + y_1^2 = r^2$. (If the point is not on the circle, there is no tangent at that point.)
Find the gradient of the radius from the origin $O(0,0)$ to $P(x_1,y_1)$: $$m_{\text{radius}} = \frac{y_1 - 0}{x_1 - 0} = \frac{y_1}{x_1}$$
Find the gradient of the tangent using $m_1 \times m_2 = -1$: $$m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{x_1}{y_1}$$
Write the equation of the tangent using $y - y_1 = m(x - x_1)$ and simplify.

Worked Example 4 — Full Method: Tangent at $(3,4)$ on $x^2+y^2=25$

①Verify: $3^2+4^2 = 9+16 = 25$ ✓ The point is on the circle.

②Gradient of radius: $m_{\text{radius}} = \dfrac{4}{3}$

③Gradient of tangent: $m_{\text{tangent}} = -\dfrac{3}{4}$ (flip and negate $\dfrac{4}{3}$)

④Equation: $y - 4 = -\dfrac{3}{4}(x - 3)$
Multiply through by 4: $4y - 16 = -3(x-3) = -3x+9$
Rearrange: $3x + 4y = 25$

✓Tangent equation: $\mathbf{3x + 4y = 25}$ (or $y = -\tfrac{3}{4}x + \tfrac{25}{4}$)

Worked Example 5 — Tangent at $(-4,3)$ on $x^2+y^2=25$

①Verify: $(-4)^2+3^2 = 16+9 = 25$ ✓

②Gradient of radius: $m_{\text{radius}} = \dfrac{3}{-4} = -\dfrac{3}{4}$

③Gradient of tangent: $m_{\text{tangent}} = -\dfrac{1}{-3/4} = \dfrac{4}{3}$

④Equation: $y - 3 = \dfrac{4}{3}(x - (-4)) = \dfrac{4}{3}(x+4)$
Multiply by 3: $3y - 9 = 4(x+4) = 4x+16$
Rearrange: $-4x + 3y = 25$, or equivalently $\mathbf{4x - 3y + 25 = 0}$

⚠️ Special cases:
• If the point is at $(r, 0)$ or $(-r, 0)$ (on the x-axis), the radius is horizontal (gradient 0), so the tangent is vertical: $x = r$ or $x = -r$.
• If the point is at $(0, r)$ or $(0, -r)$ (on the y-axis), the radius is vertical (undefined gradient), so the tangent is horizontal: $y = r$ or $y = -r$.

5 The Tangent Formula — A Useful Shortcut

⚡ Tangent Shortcut Formula

The equation of the tangent to $x^2+y^2=r^2$ at the point $(x_1,y_1)$ is:

$$\boxed{x \cdot x_1 + y \cdot y_1 = r^2}$$

Replace every $x^2$ with $x \cdot x_1$ and every $y^2$ with $y \cdot y_1$ in the circle equation — that's the tangent!

Why it works: Using the full method, the tangent at $(x_1,y_1)$ simplifies to $x \cdot x_1 + y \cdot y_1 = r^2$ in all cases. You can verify this:
At $(3,4)$ on $x^2+y^2=25$: tangent is $3x+4y=25$ — exactly $x(3)+y(4)=25$ ✓

Worked Example 6 — Using the Shortcut Formula

Find the equation of the tangent to $x^2+y^2=100$ at the point $(-6,8)$.

①Verify: $(-6)^2+8^2 = 36+64 = 100$ ✓

②Apply formula with $x_1=-6$, $y_1=8$, $r^2=100$:

③$x(-6) + y(8) = 100$

✓Tangent: $\mathbf{-6x + 8y = 100}$, which simplifies to $\mathbf{-3x+4y=50}$ (dividing by 2).

Worked Example 7 — Finding the Point of Tangency

The tangent to the circle $x^2+y^2=50$ has equation $x+7y=50$. Find the point of tangency.

①Using the formula $x\cdot x_1 + y\cdot y_1 = r^2$, the tangent at $(x_1,y_1)$ is $x\cdot x_1 + y\cdot y_1 = 50$.

②Comparing with $1\cdot x + 7 \cdot y = 50$: we need $x_1 = 1$ and $y_1 = 7$.

③Check: $1^2+7^2 = 1+49 = 50$ ✓ The point $(1,7)$ is on the circle.

✓Point of tangency: $\mathbf{(1,\ 7)}$

💡 Summary — Two ways to find the tangent equation:
Method 1 (Full): Find gradient of radius → find gradient of tangent → use $y-y_1=m(x-x_1)$.
Method 2 (Shortcut): Write $x \cdot x_1 + y \cdot y_1 = r^2$ directly. Both give the same answer — pick whichever you find easier to remember!

6 Where Does a Line Meet a Circle?

To find where a straight line meets a circle, substitute the equation of the line into the circle equation to get a quadratic. The number of solutions tells you how the line and circle relate.

Number of solutions	Relationship	Discriminant $b^2-4ac$
2 solutions	Line crosses the circle at two points (chord)	$b^2-4ac > 0$
1 solution (repeated)	Line is a tangent (touches at one point)	$b^2-4ac = 0$
No real solutions	Line misses the circle entirely	$b^2-4ac < 0$

Worked Example 8 — Finding Intersection Points

Find the coordinates where the line $y = x + 1$ meets the circle $x^2+y^2=25$.

①Substitute $y = x+1$ into $x^2+y^2=25$:

②$x^2 + (x+1)^2 = 25 \Rightarrow x^2 + x^2+2x+1 = 25 \Rightarrow 2x^2+2x-24 = 0$

③Divide by 2: $x^2+x-12 = 0 \Rightarrow (x+4)(x-3)=0$

④$x = -4$ or $x = 3$
At $x=-4$: $y=-4+1=-3$ → point $(-4,-3)$
At $x=3$: $y=3+1=4$ → point $(3,4)$

✓Two intersection points: $\mathbf{(-4,\ -3)}$ and $\mathbf{(3,\ 4)}$ — the line is a chord.

💡 Checking answers: Always substitute back: $(-4)^2+(-3)^2=16+9=25$ ✓ and $3^2+4^2=9+16=25$ ✓

7 Quick Reference

⭕ Circle equation

$x^2 + y^2 = r^2$
Centre: $(0,0)$
Radius: $r = \sqrt{r^2}$
Always positive square root

🔵 Testing a point

Substitute $(a,b)$:
$a^2+b^2 = r^2$ → on
$a^2+b^2 < r^2$ → inside
$a^2+b^2 > r^2$ → outside

📐 Perpendicular gradients

$m_1 \times m_2 = -1$
If $m_1 = \dfrac{a}{b}$, then $m_2 = -\dfrac{b}{a}$
Radius gradient $\times$ tangent gradient $= -1$

⚡ Tangent formula

At point $(x_1,y_1)$ on $x^2+y^2=r^2$:
$x\cdot x_1 + y\cdot y_1 = r^2$
Quick — no working needed!

↕ Special tangents

At $(r,0)$: tangent is $x = r$
At $(-r,0)$: tangent is $x = -r$
At $(0,r)$: tangent is $y = r$
At $(0,-r)$: tangent is $y = -r$

🔀 Line meets circle

Substitute line into circle → quadratic. Solve and use $x$-values to find coordinates. Check each answer!

Situation	What to do	Key formula/fact
Find $r$ from equation	Square-root the RHS	$r = \sqrt{r^2}$
Write equation given $r$	Square the radius	$x^2+y^2 = r^2$
Point on circle?	Substitute, compare to $r^2$	$a^2+b^2 = r^2$
Missing coordinate	Substitute known value, solve for unknown	$k^2 + b^2 = r^2$
Tangent equation	Use shortcut: $x x_1 + y y_1 = r^2$	Radius $\perp$ tangent
Point of tangency	Match coefficients in $x x_1+y y_1=r^2$	Compare to given tangent

8 Practice Questions

Question 1 — Identifying Radius

Write down the radius of each circle:

(a) $x^2+y^2=49$ (b) $x^2+y^2=3$ (c) $x^2+y^2=144$

▶ Show Solution

(a) $r = \sqrt{49} = \mathbf{7}$

(b) $r = \sqrt{3} \approx \mathbf{1.73}$

(c) $r = \sqrt{144} = \mathbf{12}$

Question 2 — Writing the Equation

Write the equation of the circle with centre the origin and:

(a) radius $11$ (b) radius $\sqrt{20}$ (c) passing through the point $(5,12)$

▶ Show Solution

(a) $r=11 \Rightarrow r^2=121 \Rightarrow$ equation: $\mathbf{x^2+y^2=121}$

(b) $r=\sqrt{20} \Rightarrow r^2=20 \Rightarrow$ equation: $\mathbf{x^2+y^2=20}$

(c) $r^2 = 5^2+12^2 = 25+144=169 \Rightarrow$ equation: $\mathbf{x^2+y^2=169}$ (radius $=13$)

Question 3 — On, Inside, or Outside?

For the circle $x^2+y^2=50$, state whether each point is on, inside, or outside the circle. Show your working.

(a) $(5,5)$ (b) $(6,4)$ (c) $(7,2)$ (d) $(-7,1)$

▶ Show Solution

(a) $5^2+5^2=25+25=50=50$ → On the circle

(b) $6^2+4^2=36+16=52 > 50$ → Outside the circle

(c) $7^2+2^2=49+4=53 > 50$ → Outside the circle

(d) $(-7)^2+1^2=49+1=50=50$ → On the circle

Question 4 — Missing Coordinate

The point $(3,k)$ lies on the circle $x^2+y^2=58$. Find all possible values of $k$.

▶ Show Solution

Substitute: $3^2+k^2=58 \Rightarrow 9+k^2=58 \Rightarrow k^2=49 \Rightarrow k=\pm 7$

The two points are $(3,7)$ and $(3,-7)$. Check: $3^2+7^2=9+49=58$ ✓

Question 5 — Tangent at a Given Point (Full Method)

Find the equation of the tangent to the circle $x^2+y^2=25$ at the point $(4,-3)$. Give your answer in the form $ax+by=c$.

▶ Show Solution

Verify: $4^2+(-3)^2=16+9=25$ ✓

Gradient of radius: $m_{\text{radius}} = \dfrac{-3}{4}$

Gradient of tangent: $m_{\text{tangent}} = -\dfrac{1}{-3/4} = \dfrac{4}{3}$

Equation: $y-(-3) = \dfrac{4}{3}(x-4)$
$3(y+3) = 4(x-4)$
$3y+9 = 4x-16$
$\mathbf{4x-3y=25}$

Or using shortcut: $4x+(-3)y=25 \Rightarrow 4x-3y=25$ ✓

Question 6 — Tangent Using the Shortcut Formula

(a) Show that the point $A(5,12)$ lies on the circle $x^2+y^2=169$.

(b) Find the equation of the tangent to the circle at $A$.

▶ Show Solution

(a) $5^2+12^2=25+144=169=169$ ✓ So $A$ lies on the circle.

(b) Using the shortcut: tangent at $(5,12)$ on $x^2+y^2=169$:
$x(5)+y(12)=169$
$\mathbf{5x+12y=169}$

Question 7 — Tangent at a Point on the Axes

A circle has equation $x^2+y^2=36$. Write down the equation of the tangent to the circle at:

(a) the point $(6,0)$ (b) the point $(0,-6)$ (c) the point $(-6,0)$

▶ Show Solution

(a) At $(6,0)$: the radius is horizontal, so the tangent is vertical: $\mathbf{x=6}$

Or using shortcut: $x(6)+y(0)=36 \Rightarrow 6x=36 \Rightarrow x=6$ ✓

(b) At $(0,-6)$: the radius is vertical, so the tangent is horizontal: $\mathbf{y=-6}$

(c) At $(-6,0)$: the radius is horizontal, so the tangent is vertical: $\mathbf{x=-6}$

Question 8 — Finding the Point of Tangency

The line $7x+9y=130$ is a tangent to the circle $x^2+y^2=130$.

(a) Find the coordinates of the point of tangency.

(b) Verify that this point lies on the circle.

▶ Show Solution

(a) The tangent at $(x_1,y_1)$ on $x^2+y^2=130$ has equation $x\cdot x_1+y\cdot y_1=130$.
Comparing with $7x+9y=130$: we need $x_1=7$ and $y_1=9$.
Point of tangency: $\mathbf{(7,9)}$

(b) Check: $7^2+9^2=49+81=130$ ✓ The point lies on the circle.

Question 9 — Line Meets Circle

Find the coordinates of the points where the line $y = 2x - 5$ meets the circle $x^2+y^2=20$.

▶ Show Solution

Substitute $y=2x-5$ into $x^2+y^2=20$:

$x^2+(2x-5)^2=20$

$x^2+4x^2-20x+25=20$

$5x^2-20x+5=0$

$x^2-4x+1=0$ (dividing by 5)

$x = \dfrac{4 \pm \sqrt{16-4}}{2} = \dfrac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$

At $x=2+\sqrt{3}$: $y=2(2+\sqrt{3})-5=4+2\sqrt{3}-5=-1+2\sqrt{3}$

At $x=2-\sqrt{3}$: $y=-1-2\sqrt{3}$

Points: $\mathbf{(2+\sqrt{3},\ -1+2\sqrt{3})}$ and $\mathbf{(2-\sqrt{3},\ -1-2\sqrt{3})}$

Question 10 — Multi-Step Problem

The point $Q(-3,k)$ lies on the circle $x^2+y^2=25$, where $k > 0$.

(a) Find the value of $k$.

(b) Find the gradient of the radius $OQ$.

(d) Find the coordinates of where the tangent crosses the $x$-axis and the $y$-axis.

▶ Show Solution

(a) $(-3)^2+k^2=25 \Rightarrow 9+k^2=25 \Rightarrow k^2=16 \Rightarrow k=4$ (since $k>0$)
So $Q=(-3,4)$.

(b) Gradient of $OQ = \dfrac{4-0}{-3-0} = -\dfrac{4}{3}$

(c) Using shortcut: tangent at $(-3,4)$: $x(-3)+y(4)=25$
$\mathbf{-3x+4y=25}$ (or equivalently $3x-4y+25=0$)

(d) x-intercept (set $y=0$): $-3x=25 \Rightarrow x=-\dfrac{25}{3}$ → crosses x-axis at $\left(-\dfrac{25}{3}, 0\right)$
y-intercept (set $x=0$): $4y=25 \Rightarrow y=\dfrac{25}{4}$ → crosses y-axis at $\left(0, \dfrac{25}{4}\right)$