Circles in coordinate geometry

A circle is the set of all points a fixed distance from a centre. Coordinate geometry translates that definition into an equation, allowing geometric facts about radii, chords and tangents to be combined with algebra.

Before you begin

You should be confident with:

• the distance and midpoint formulae
• straight-line gradients and equations
• expanding brackets and completing the square
• solving simultaneous equations

The equation of a circle

Let the centre be $C(a,b)$ and the radius be $r$. A general point $P(x,y)$ lies on the circle exactly when $CP=r$.

Using the distance formula,

$\sqrt{(x-a)^2+(y-b)^2}=r.$

Squaring gives the standard form

$\boxed{(x-a)^2+(y-b)^2=r^2}.$

Read the centre by reversing the signs inside the brackets:

$(x-3)^2+(y+4)^2=25$

has centre $(3,-4)$ and radius $5$.

The right-hand side is $r^2$, not $r$. For

$x^2+y^2=13,$

the centre is $(0,0)$ and the radius is $\sqrt{13}$.

Example 1: write an equation from a centre and point

A circle has centre $(-2,3)$ and passes through $(4,-5)$. Find its equation.

First calculate the squared radius:

$r^2=(4-(-2))^2+(-5-3)^2=6^2+(-8)^2=100.$

Therefore

$\boxed{(x+2)^2+(y-3)^2=100}.$

Using $r^2$ directly avoids taking a square root only to square it again.

Expanded form and completing the square

Expanding the standard equation produces

$x^2+y^2-2ax-2by+(a^2+b^2-r^2)=0.$

Consequently, an equation of a circle has:

• equal coefficients of $x^2$ and $y^2$, after any common factor is removed
• no $xy$ term

An equation such as $x^2+2y^2=8$ describes an ellipse, not a circle.

To recover the centre and radius from an expanded equation, complete the square separately in $x$ and $y$:

$x^2+px=\left(x+\frac p2\right)^2-\left(\frac p2\right)^2.$

Example 2: find the centre and radius

Find the centre and radius of

$x^2+y^2-6x+8y-11=0.$

Group the variable terms and move the constant:

$x^2-6x+y^2+8y=11.$

Complete both squares:

$x^2-6x=(x-3)^2-9,$

$y^2+8y=(y+4)^2-16.$

Hence

$(x-3)^2-9+(y+4)^2-16=11,$

so

$\boxed{(x-3)^2+(y+4)^2=36}.$

The centre is $(3,-4)$ and the radius is $6$.

Example 3: a common coefficient

Find the centre and radius of

$2x^2+2y^2+12x-4y-30=0.$

First divide every term by $2$:

$x^2+y^2+6x-2y-15=0.$

Then complete the squares:

$(x+3)^2-9+(y-1)^2-1=15,$

$\boxed{(x+3)^2+(y-1)^2=25}.$

The centre is $(-3,1)$ and the radius is $5$.

Does a point lie on the circle?

For a circle with centre $(a,b)$ and radius $r$, compare

$D=(x-a)^2+(y-b)^2$

with $r^2$.

Comparison	Position of the point
$D<r^2$	Inside the circle
$D=r^2$	On the circle
$D>r^2$	Outside the circle

For the circle

$(x-2)^2+(y+1)^2=20,$

consider $P(4,3)$. Substitution gives

$D=(4-2)^2+(3+1)^2=4+16=20.$

Therefore $P$ lies on the circle.

This test is also a quick way to check a coordinate obtained from longer algebra.

Intersections of a line and a circle

A line can meet a circle at two points, touch it once, or miss it. Substitute the line equation into the circle equation and inspect the resulting quadratic.

Discriminant	Geometric result
$b^2-4ac>0$	Two intersections
$b^2-4ac=0$	One repeated root, so the line is tangent
$b^2-4ac<0$	No real intersection

Example 4: find line-circle intersections

Find where

$x^2+y^2=25$

meets the line $y=x+1$.

Substitute $y=x+1$:

$x^2+(x+1)^2=25,$

$2x^2+2x-24=0,$

$x^2+x-12=0,$

$(x+4)(x-3)=0.$

Thus $x=-4$ or $x=3$. Since $y=x+1$, the intersection points are

$\boxed{(-4,-3) \text{ and } (3,4)}.$

Always return coordinates, not just the solved $x$ values.

Radius and tangent are perpendicular

At a point $P$ on a circle, the tangent is perpendicular to the radius $CP$.

To see why, take any other point $Q$ on the tangent. The perpendicular distance from $C$ to the tangent is $CP$. Every other segment $CQ$ is longer than $CP$, so $Q$ lies outside the circle. Therefore the line meets the circle only at $P$, exactly as a tangent should.

This gives a direct coordinate method:

1. Find the gradient of the radius from the centre to the point of contact.
2. Take its negative reciprocal to obtain the tangent gradient.
3. Use the point of contact in point-gradient form.

Example 5: find a tangent and normal

Find the tangent and normal to

$(x-1)^2+(y+2)^2=25$

at $P(4,2)$.

The centre is $C(1,-2)$. Check the point:

$(4-1)^2+(2+2)^2=9+16=25.$

The radius gradient is

$m_{CP}=\frac{2-(-2)}{4-1}=\frac43.$

The tangent gradient is its negative reciprocal, $-3/4$. Therefore

$\boxed{y-2=-\frac34(x-4)}.$

The normal lies along the radius, so

$\boxed{y-2=\frac43(x-4)}.$

The normal to a circle always passes through its centre.

Horizontal and vertical radii

If the radius is horizontal, its gradient is $0$ and the tangent is vertical. If the radius is vertical, the tangent is horizontal.

For the circle $(x-2)^2+(y-3)^2=16$:

• at $(6,3)$, the tangent is $x=6$
• at $(2,7)$, the tangent is $y=7$

Do not try to take a numerical negative reciprocal of zero or of an undefined gradient.

A direct tangent formula

For the circle

$(x-a)^2+(y-b)^2=r^2,$

the tangent at a point $(x_1,y_1)$ on the circle is

$\boxed{(x_1-a)(x-a)+(y_1-b)(y-b)=r^2}.$

This follows because the vectors

\quad\text{and}\quad \begin{pmatrix}x_1-a\\y_1-b\end{pmatrix}$$ have dot product $r^2$ precisely along the tangent through $(x_1,y_1)$. It is a useful shortcut, but the radius-gradient method is easier to reconstruct under pressure. For the circle $x^2+y^2=25$ at $(3,4)$, the formula gives $$\boxed{3x+4y=25}.$$ The normal is the line through $(0,0)$ and $(3,4)$, namely $y=4x/3$. For more general curve methods, see [tangents and normals](/learn/calculus/tangents-and-normals/). ## The centre-to-chord property A chord joins two points on a circle. The perpendicular from the centre to a chord bisects the chord. Suppose $AB$ is a chord and the perpendicular from centre $O$ meets it at $M$. The right-angled triangles $OMA$ and $OMB$ have: - $OA=OB$, since both are radii - $OM$ in common - right angles at $M$ The triangles are congruent by RHS, so $AM=MB$. The converse is also useful: the line from the centre to the midpoint of a chord is perpendicular to the chord. ### Example 6: length of a chord A circle has radius $13$. A chord is $5$ units from the centre. Find its length. The perpendicular from the centre bisects the chord. If half the chord has length $q$, Pythagoras gives $$q^2+5^2=13^2.$$ Therefore $$q=\sqrt{169-25}=12.$$ The whole chord has length $$\boxed{24}.$$ ### Example 7: equation of a chord's perpendicular bisector The endpoints of a chord are $A(-1,2)$ and $B(5,6)$. A circle through them has its centre on the chord's perpendicular bisector. The midpoint is $$M=\left(\frac{-1+5}{2},\frac{2+6}{2}\right)=(2,4).$$ The chord gradient is $$\frac{6-2}{5-(-1)}=\frac23,$$ so the perpendicular gradient is $-3/2$. The perpendicular bisector is $$\boxed{y-4=-\frac32(x-2)}.$$ Two different chords give two perpendicular bisectors, whose intersection is the circle's centre. ## The angle in a semicircle If $AB$ is a diameter and $P$ is any other point on the circle, then $$\boxed{\angle APB=90^\circ}.$$ This is sometimes called Thales' theorem. ### Why it is true Join the centre $O$ to $P$. Let $$\angle OAP=\alpha, \qquad \angle OBP=\beta.$$ Since $OA=OP$, triangle $AOP$ is isosceles, so $\angle APO=\alpha$. Similarly, $OB=OP$ gives $\angle OPB=\beta$. The angle at $P$ is therefore $$\angle APB=\alpha+\beta.$$ In triangle $APB$, the other two angles are also $\alpha$ and $\beta$, because $A$, $O$ and $B$ are collinear. Hence $$2\alpha+2\beta=180^\circ,$$ so $$\alpha+\beta=90^\circ.$$ Therefore $\angle APB=90^\circ$. ### Example 8: equation from a diameter The endpoints of a diameter are $A(-2,1)$ and $B(6,7)$. Find the circle's equation. The centre is the midpoint: $$C=\left(\frac{-2+6}{2},\frac{1+7}{2}\right)=(2,4).$$ The squared diameter is $$AB^2=(6-(-2))^2+(7-1)^2=8^2+6^2=100.$$ The radius is half the diameter, so $r^2=100/4=25$. Therefore $$\boxed{(x-2)^2+(y-4)^2=25}.$$ Any third point $P$ on this circle forms a right-angled triangle $APB$. ## Finding a circle from three points Three non-collinear points determine one circle. There are two common approaches: 1. Find two perpendicular bisectors and intersect them to locate the centre. 2. Substitute all three points into $x^2+y^2+Dx+Ey+F=0$ and solve for $D$, $E$ and $F$. The geometric method explains the construction. The algebraic method is often faster with awkward coordinates. ### Example 9: use the expanded equation Find the circle through $(0,0)$, $(4,0)$ and $(0,6)$. Start with $$x^2+y^2+Dx+Ey+F=0.$$ Substituting $(0,0)$ gives $F=0$. Substituting $(4,0)$ gives $$16+4D=0 \quad\Rightarrow\quad D=-4.$$ Substituting $(0,6)$ gives $$36+6E=0 \quad\Rightarrow\quad E=-6.$$ Thus $$x^2+y^2-4x-6y=0.$$ Complete the square: $$\boxed{(x-2)^2+(y-3)^2=13}.$$ The points form a right-angled triangle at the origin, so the angle-in-a-semicircle theorem also tells us that the hypotenuse from $(4,0)$ to $(0,6)$ is a diameter. Its midpoint $(2,3)$ confirms the centre. ## Common mistakes - Reading the centre of $(x+4)^2+(y-2)^2=9$ as $(4,-2)$ instead of $(-4,2)$. - Calling the right-hand side the radius instead of the squared radius. - Completing the square in $x$ but forgetting to do the same independently in $y$. - Adding a completing-square correction on only one side of an equation. - Finding $x$ coordinates of intersections but not the corresponding $y$ coordinates. - Assuming a point lies on a circle without checking it. - Using the radius gradient as the tangent gradient instead of taking the negative reciprocal. - Forgetting that vertical lines have equations of the form $x=c$. - Using the full chord length in the right-angled triangle instead of half the chord. ## Check your understanding 1. Find the centre and radius of $x^2+y^2+10x-4y-20=0$. 2. Determine whether $(5,1)$ is inside, on or outside the circle $(x-1)^2+(y+2)^2=25$. 3. Find the tangent and normal to $x^2+y^2=34$ at $(3,5)$. 4. A circle has centre $(2,-1)$ and the line $x=8$ is tangent to it. Find its equation. 5. A chord of a circle has length $16$ and is $6$ units from the centre. Find the radius. <details> <summary>Answers</summary> ### 1 $$x^2+10x+y^2-4y=20,$$ $$ (x+5)^2-25+(y-2)^2-4=20,$$ $$\boxed{(x+5)^2+(y-2)^2=49}.$$ The centre is $(-5,2)$ and the radius is $7$. ### 2 $$D=(5-1)^2+(1+2)^2=16+9=25.$$ The point lies on the circle. ### 3 The radius from $(0,0)$ to $(3,5)$ has gradient $5/3$. The tangent gradient is $-3/5$. Therefore $$\boxed{y-5=-\frac35(x-3)}$$ and the normal is $$\boxed{y-5=\frac53(x-3)}.$$ ### 4 The perpendicular distance from $(2,-1)$ to $x=8$ is $6$, so the radius is $6$: $$\boxed{(x-2)^2+(y+1)^2=36}.$$ ### 5 Half the chord is $8$. By Pythagoras, $$r^2=8^2+6^2=100,$$ so $\boxed{r=10}$. </details> ## What to learn next Use [straight lines](/learn/coordinate-geometry/straight-lines/) to strengthen gradient and perpendicular-bisector methods. Continue to [tangents and normals](/learn/calculus/tangents-and-normals/) for general curves, or explore the [interactive circumference model](/learn/lab/circle-circumference/) to connect radius, diameter and circumference visually.