Circle

Circle Calculator

Choose what you know about the circle (its radius, diameter, circumference or area) and type the value. The calculator finds the other three step by step: first the exact value with $\pi,$ then a rounded value. A circumference or an area can be typed as a multiple of $\pi\text{:}$ write pi, for example 10pi. If you also type a central angle in degrees, the calculator finds the length of its arc and the area of its sector. It uses the exact value of $\pi,$ so its last digit can differ from a calculation with ${\pi\approx 3.14}.$

What Is a Circle?

A circle is the set of all points in a plane that are at the same distance from a given point $O,$ the center of the circle. This distance is the radius $r$ of the circle.

The circle is only the round line. The circle together with all the points inside it is called a disk. When we speak of the area of a circle, we mean the area of this disk: a circle as a line has a length, the circumference, but no area. A point $X$ lies inside the circle if ${OX\lt r},$ on the circle if ${OX=r},$ and outside the circle if ${OX\gt r}.$

On this page: Calculator · Parts of a circle · Circumference and area · Arcs and sectors · Angles in a circle · Chords and tangents · Solved problems · Practice

Parts of a Circle

In the figure $O$ is the center of the circle.

• A radius is a segment that joins the center with a point of the circle, here $OA.$ All radii of a circle are equal, and their length is also called the radius $r.$
• A chord is a segment that joins two points of the circle, here $DE.$
• A diameter is a chord that passes through the center, here $BC.$ It consists of two radii, so its length is ${d=2r}.$ The diameter is the longest chord of the circle: by the triangle inequality, any chord $DE$ is at most ${DO+OE=2r}.$
• An arc is a part of the circle between two of its points, here the arc $FG.$ We write it as $\overset{\frown}{FG}.$ Two points divide a circle into two arcs; a diameter divides it into two semicircles.

A line and a circle can have two, one or no common points. Which case occurs depends on the distance from the center to the line: it is less than $r,$ equal to $r$ or greater than $r.$ A line that crosses a circle at two points is a secant, like the line $MN.$ A line that has exactly one common point with the circle is a tangent, and this point $T$ is the point of tangency. The tangent is perpendicular to the radius $OT$ (see Tangents below).

Two radii divide the disk into two sectors, shaped like slices of a pie. A chord divides the disk into two segments: a segment is the part of the disk between a chord and an arc.

Circumference and Area of a Circle

The Number $\pi$

The length of a circle is called its circumference $C.$ In every circle, small or large, the circumference is a little more than 3 times the diameter, and the ratio $\frac{C}{d}$ is always the same number. This number is called $\pi$ (pi):

A wheel with a diameter of 1 rolls exactly $\pi$ units in one full turn. The decimal digits of $\pi$ never end and never repeat. In calculations we keep $\pi$ in the answer or use an approximate value, $3.14$ or $\frac{22}{7}.$

Circumference

Example. A circle has a radius of 5 cm. Its circumference is ${C=2\pi\cdot 5=10\pi}$ cm. With ${\pi\approx 3.14}$ this is about $31.4$ cm.

From the circumference we can find the radius: ${r=\frac{C}{2\pi}}.$ For example, if ${C=18.84}$ cm, then ${r\approx\frac{18.84}{2\cdot 3.14}=3}$ cm.

Area of a Circle

Why? Cut the disk into many equal sectors and put them side by side, turning every second one upside down. They form a figure that is almost a parallelogram. Its height is the radius $r,$ and its base is half the circumference, ${\frac{C}{2}=\pi r}.$ The more sectors we take, the closer the figure gets to a rectangle with the sides $\pi r$ and $r,$ so the area of the disk is ${\pi r\cdot r=\pi r^2}.$

Example. The area of a disk with a radius of 5 cm is ${A=\pi\cdot 5^2=25\pi}$ cm², about $78.5$ cm². With the diameter: since ${r=\frac{d}{2}},$ the area is also ${A=\frac{\pi d^2}{4}}.$

From the area we can find the radius: ${r=\sqrt{\frac{A}{\pi}}}.$ For example, if ${A=49\pi}$ cm², then ${r=\sqrt{49}=7}$ cm.

Arcs and Sectors

Central Angle and the Measure of an Arc

An angle with its vertex at the center of a circle is a central angle. In the figure ${\angle AOB=\alpha}$ is a central angle. The arc $AB$ inside it is the arc that the angle intercepts.

An arc is measured in degrees by its central angle: ${\overset{\frown}{AB}=\angle AOB}.$

The whole circle is $360^\circ,$ a semicircle is $180^\circ,$ and a quarter of a circle is $90^\circ.$ An arc smaller than a semicircle is a minor arc, and an arc larger than a semicircle is a major arc. If the minor arc $AB$ is $80^\circ,$ the major arc with the same ends is ${360^\circ-80^\circ=280^\circ}.$

Arc Length

An arc of $\alpha^\circ$ is $\frac{\alpha}{360}$ of the whole circle, so its length is $\frac{\alpha}{360}$ of the circumference:

Example. In a circle with a radius of 6 cm, the arc of $60^\circ$ is ${\ell=\frac{\pi\cdot 6\cdot 60}{180}=2\pi}$ cm, about $6.28$ cm.

If we know the length of the arc, we find its central angle from the same formula: ${\alpha=\frac{180\,\ell}{\pi r}}.$

Area of a Sector

A sector with a central angle of $\alpha^\circ$ is $\frac{\alpha}{360}$ of the disk, so its area is:

Example. A sector of $90^\circ$ in a circle with a radius of 4 cm is a quarter of the disk: ${A_{\text{sector}}}={\frac{90}{360}\cdot\pi\cdot 4^2}={4\pi}$ cm², about $12.57$ cm².

Angles can also be measured in radians: a central angle of $1$ radian intercepts an arc as long as the radius, and ${180^\circ=\pi}$ radians. If $\alpha$ is in radians, the formulas are simpler: ${\ell=r\alpha}$ and ${A_{\text{sector}}=\frac{r^2\alpha}{2}}.$

Area of a Segment

The segment cut off by a chord $AB$ (the smaller one) is the sector $AOB$ without the triangle $AOB\text{:}$

For the larger segment, add the area of the triangle to the area of the larger sector instead. See Problem 5 in the solved problems below.

Area of a Ring

A ring (an annulus) is the region between two circles with a common center and the radii $R$ and ${r\lt R}.$ Its area is the area of the large disk minus the area of the small one:

Area of a Part of a Ring

Two radii with a central angle $\alpha^\circ$ cut a part off the ring (on the right in the figure above; it is called an annular sector). It is $\frac{\alpha}{360}$ of the ring:

Angles in a Circle

Inscribed Angle

An angle whose vertex lies on the circle and whose sides are chords is an inscribed angle. In the figure $\angle APB$ is an inscribed angle. It intercepts the arc $AB$ that lies inside it (the arc without the point $P$), and $\angle AOB$ is the central angle on the same arc.

Inscribed Angle Theorem. An inscribed angle is half the arc it intercepts: $$\angle APB=\frac{\overset{\frown}{AB}}{2}.$$ When this arc is a minor arc, as in the figure, the inscribed angle is also half the central angle: ${\angle APB=\frac{\angle AOB}{2}}.$

Proof

1. One side of the angle is a diameter. Let $PA$ pass through the center $O.$ The triangle $OPB$ is isosceles, because ${OP=OB=r},$ so ${\angle OBP=\angle OPB=\angle APB}.$ The central angle $AOB$ is an exterior angle of the triangle $OPB,$ so it equals the sum of the two angles not adjacent to it: ${\angle AOB}={\angle OPB+\angle OBP}={2\angle APB}.$ So ${\angle APB}={\frac{\angle AOB}{2}}={\frac{\overset{\frown}{AB}}{2}}.$

2. The center lies inside the angle. Draw the diameter $PC.$ It divides the angle and the arc into two parts, and case 1 applies to each part: ${\angle APB}={\angle APC+\angle CPB}={\frac{\overset{\frown}{AC}}{2}+\frac{\overset{\frown}{CB}}{2}}={\frac{\overset{\frown}{AB}}{2}}.$

3. The center lies outside the angle. Draw the diameter $PC$ again. Now the angle is a difference: ${\angle APB}={\angle CPB-\angle CPA}={\frac{\overset{\frown}{CB}}{2}-\frac{\overset{\frown}{CA}}{2}}={\frac{\overset{\frown}{AB}}{2}}.$

Example. The arc $AB$ measures $84^\circ.$ Then the inscribed angle $APB$ is ${\frac{84^\circ}{2}=42^\circ}.$

Two important consequences:

• Inscribed angles that intercept the same arc are equal: in the figure on the left, ${\angle AP_1B=\angle AP_2B}.$
• An angle inscribed in a semicircle is a right angle: if $AB$ is a diameter, then ${\angle APB=\frac{180^\circ}{2}=90^\circ}$ for every point $P$ of the circle other than $A$ and $B$ (in the figure on the right).

Angle Between a Tangent and a Chord

The angle between a tangent and a chord drawn from the point of tangency is half the arc inside this angle: ${\angle KAB=\frac{\overset{\frown}{AB}}{2}}.$

Proof

Draw the diameter $AC.$ The tangent is perpendicular to the radius (see Tangents), so ${\angle KAC=90^\circ}$ and ${\angle KAB=90^\circ-\angle BAC}.$ The angle $ABC$ is inscribed in a semicircle, so it is a right angle, and in the triangle $ABC$ we have ${\angle ACB=90^\circ-\angle BAC}.$ Hence ${\angle KAB=\angle ACB},$ and the inscribed angle $ACB$ is half the arc $AB.$

If the angle $KAB$ is obtuse, the other angle between the chord and the tangent is acute and is half the other arc, so ${\angle KAB}={180^\circ-\frac{360^\circ-\overset{\frown}{AB}}{2}}={\frac{\overset{\frown}{AB}}{2}}.$

Angles Between Two Chords or Secants

Case 1: two chords meet inside the circle.

The angle between two chords that meet inside a circle is half the sum of the arcs intercepted by this angle and by its vertical angle: $$\angle AEB=\frac{\overset{\frown}{AB}+\overset{\frown}{CD}}{2}.$$

Proof

Draw the segment $BC.$ The angle $AEB$ is an exterior angle of the triangle $BEC,$ so ${\angle AEB=\angle EBC+\angle ECB}.$ The point $E$ lies on $BD$ and on $AC,$ so these are the inscribed angles ${\angle EBC=\angle DBC=\frac{\overset{\frown}{CD}}{2}}$ and ${\angle ECB=\angle ACB=\frac{\overset{\frown}{AB}}{2}}.$ Adding them gives the formula.

Example. In the figure the arc $AB$ is $60^\circ$ and the arc $CD$ is $50^\circ.$ Then the angles 1 and 2 (they are vertical angles) are both ${\frac{60^\circ+50^\circ}{2}=55^\circ}.$

Case 2: two secants meet outside the circle.

The angle between two secants drawn from a point outside a circle is half the difference of the intercepted arcs, the far one minus the near one: $$\angle BPD=\frac{\overset{\frown}{BD}-\overset{\frown}{AC}}{2}.$$

Proof

Draw the segment $AD.$ The angle $BAD$ is an exterior angle of the triangle $APD,$ so ${\angle BAD=\angle APD+\angle ADP}.$ The angles $BAD$ and $ADP$ are inscribed angles (the point $C$ lies on $DP$): ${\angle BAD=\frac{\overset{\frown}{BD}}{2}}$ and ${\angle ADP=\angle ADC=\frac{\overset{\frown}{AC}}{2}}.$ The point $A$ lies on $PB,$ so ${\angle BPD}={\angle APD}={\frac{\overset{\frown}{BD}}{2}-\frac{\overset{\frown}{AC}}{2}}.$

Example. If the far arc measures $80^\circ$ and the near arc $30^\circ,$ the angle between the secants is ${\frac{80^\circ-30^\circ}{2}=25^\circ}.$ The same formula holds when one or both of the lines are tangents.

Chords, Tangents and Secants

A Chord and the Center

The perpendicular from the center of a circle to a chord bisects the chord: if ${OM\perp AB},$ then ${AM=MB}.$ The distance $OM$ from the center to the chord satisfies $$OM^2+\left(\frac{AB}{2}\right)^2=r^2.$$

Proof

The right triangles $OMA$ and $OMB$ have the common leg $OM$ and the equal hypotenuses ${OA=OB=r},$ so they are congruent, and ${AM=MB}.$ The second formula is the Pythagorean theorem in the triangle $OMA.$

So equal chords are at equal distances from the center, and the closer a chord is to the center, the longer it is.

Tangents

Proof

1. Every point of the tangent other than $A$ lies outside the circle (a line through a point inside the circle would cross it twice), so it is farther from $O$ than $A.$ So $OA$ is the shortest segment from $O$ to the tangent, and the shortest segment from a point to a line is the perpendicular.

2. The right triangles $OAP$ and $OBP$ have the common hypotenuse $OP$ and the equal legs ${OA=OB=r},$ so they are congruent. Hence ${PA=PB}$ and ${\angle APO=\angle BPO}.$

Intersecting Chords

If two chords $AB$ and $CD$ meet at a point $X,$ then $$AX\cdot XB=CX\cdot XD.$$

Proof

In the triangles $AXC$ and $DXB$ the angles at $X$ are vertical angles, and ${\angle CAB=\angle CDB},$ because they are inscribed angles on the same arc $CB.$ So the triangles are similar, and ${\frac{AX}{DX}=\frac{CX}{BX}},$ that is ${AX\cdot XB=CX\cdot XD}.$

Secants and Tangents from One Point

Proof

1. The triangles $PAD$ and $PCB$ have the common angle $P,$ and ${\angle PDA=\angle PBC},$ because they are inscribed angles on the same arc $AC.$ So the triangles are similar: ${\frac{PA}{PC}=\frac{PD}{PB}},$ that is ${PA\cdot PB=PC\cdot PD}.$

2. The triangles $PTA$ and $PBT$ have the common angle $P,$ and the angle between the tangent and the chord, ${\angle PTA},$ equals the inscribed angle ${\angle TBA}$ on the same arc $TA.$ So the triangles are similar: ${\frac{PT}{PB}=\frac{PA}{PT}},$ that is ${PT^2=PA\cdot PB}.$

Solved Problems

Problem 1. The radius of a circle is 7 cm. Find its diameter, circumference and area.

Solution. ${d=2\cdot 7=14}$ cm, ${C}={2\pi\cdot 7}={14\pi}\approx{43.98}$ cm and ${A}={\pi\cdot 7^2}={49\pi}\approx{153.94}$ cm².

Problem 2. The circumference of a circle is $31.4$ cm. Find its radius and its area. Take ${\pi\approx 3.14}.$

Solution. ${r=\frac{C}{2\pi}\approx\frac{31.4}{6.28}=5}$ cm, and ${A}={\pi r^2}\approx{3.14\cdot 25}={78.5}$ cm².

Problem 3. The wheel of a bicycle has a diameter of 70 cm. How many full turns does it make on a 1 km ride?

Solution. In one turn the bicycle moves forward by the circumference of the wheel: ${C=70\pi\approx 219.91}$ cm. We have ${1\text{ km}=100{,}000}$ cm, and ${\frac{100{,}000}{219.91}\approx 454.7}.$ So the wheel makes 454 full turns.

Problem 4. The radius of a circle is 6 cm. Find the length of an arc of $120^\circ$ and the area of the sector with this arc.

Solution. The arc is ${\frac{120}{360}=\frac{1}{3}}$ of the circle. So ${\ell}={\frac{1}{3}\cdot 2\pi\cdot 6}={4\pi}\approx{12.57}$ cm, and the area of the sector is ${\frac{1}{3}\cdot\pi\cdot 6^2=12\pi\approx 37.70}$ cm².

Problem 5. In a circle with the center $O$ and a radius of 10 cm, the chord $AB$ subtends the central angle ${\angle AOB=90^\circ}.$ Find the area of the smaller segment cut off by this chord.

Solution. The sector $AOB$ is a quarter of the disk: its area is ${\frac{\pi\cdot 10^2}{4}=25\pi}$ cm². The triangle $AOB$ is a right triangle with legs of 10 cm, so its area is ${\frac{10\cdot 10}{2}=50}$ cm². The area of the segment is ${25\pi-50\approx 28.54}$ cm².

Problem 6. The points $A,$ $B$ and $C$ lie on a circle with the center $O,$ and ${\angle AOB=110^\circ}.$ Find the angle $ACB$ if $C$ lies a) on the major arc $AB;$ b) on the minor arc $AB.$

Solution. a) The angle $ACB$ intercepts the minor arc $AB$ of $110^\circ,$ so ${\angle ACB=55^\circ}.$ b) Now it intercepts the major arc of ${360^\circ-110^\circ=250^\circ},$ so ${\angle ACB=125^\circ}.$ Note that ${55^\circ+125^\circ=180^\circ}:$ the two arcs make up the whole circle, so the two angles make up half of $360^\circ.$ For the same reason, the opposite angles of a quadrilateral inscribed in a circle add up to $180^\circ.$

Problem 7. A chord of 24 cm is 5 cm away from the center of a circle. Find the radius.

Solution. The perpendicular from the center bisects the chord, so each half is 12 cm. By the Pythagorean theorem ${r}={\sqrt{12^2+5^2}}={\sqrt{169}}={13}$ cm.

Problem 8. The chords $AB$ and $CD$ meet at the point $X.$ ${AX=4}$ cm, ${XB=6}$ cm and ${CX=3}$ cm. Find the length of the chord $CD.$

Solution. From ${AX\cdot XB=CX\cdot XD}$ we get ${4\cdot 6=3\cdot XD},$ so ${XD=8}$ cm and ${CD=3+8=11}$ cm.

Problem 9. The point $P$ is 13 cm away from the center of a circle with a radius of 5 cm. Find the length of the tangent from $P$ to the circle.

Solution. The tangent $PT$ is perpendicular to the radius $OT,$ so the triangle $OTP$ has a right angle at $T.$ Then ${PT}={\sqrt{13^2-5^2}}={\sqrt{144}}={12}$ cm.

Practice Problems

1. The diameter of a circle is 10 cm. Find its circumference and area.

${C=10\pi\approx 31.42}$ cm, ${A=25\pi\approx 78.54}$ cm².

2. The area of a disk is 154 cm². Find its radius and circumference. Take ${\pi\approx\frac{22}{7}}.$

${r^2\approx 154\cdot\frac{7}{22}=49},$ so ${r=7}$ cm and ${C\approx 2\cdot\frac{22}{7}\cdot 7=44}$ cm.

3. The circumference of a circle is $18\pi$ cm. Find its area.

${r=\frac{18\pi}{2\pi}=9}$ cm, so ${A=81\pi\approx 254.47}$ cm².

4. The radius of a circle is doubled. How do its circumference and area change?

The circumference doubles, and the area becomes 4 times larger: ${\pi(2r)^2=4\pi r^2}.$

5. In a circle with a radius of 9 cm, an arc is $5\pi$ cm long. Find its central angle.

${\alpha=\frac{180\cdot 5\pi}{\pi\cdot 9}=100^\circ}.$

6. A sector of a disk with a radius of 6 cm has an area of $6\pi$ cm². Find its central angle.

The sector is ${\frac{6\pi}{36\pi}=\frac{1}{6}}$ of the disk, so ${\alpha=\frac{360^\circ}{6}=60^\circ}.$

7. Find the area of a ring between two circles with a common center and the radii 5 cm and 3 cm.

${A}={\pi\left(5^2-3^2\right)}={16\pi}\approx{50.27}$ cm².

8. An inscribed angle is $38^\circ.$ Find the arc it intercepts and the central angle on the same arc.

The arc and the central angle are both ${2\cdot 38^\circ=76^\circ}.$

9. $AB$ is a diameter of a circle, ${AB=10}$ cm, and $C$ is a point of the circle with ${AC=6}$ cm. Find $BC.$

The angle $ACB$ is inscribed in a semicircle, so ${\angle ACB=90^\circ}$ and ${BC=\sqrt{10^2-6^2}=8}$ cm.

10. Two chords meet inside a circle. The arcs intercepted by one of the angles between them and by its vertical angle are $70^\circ$ and $110^\circ.$ Find this angle.

${\frac{70^\circ+110^\circ}{2}=90^\circ}:$ the chords are perpendicular.

11. Two secants are drawn from a point outside a circle. The arcs between them are $130^\circ$ and $40^\circ.$ Find the angle between the secants.

${\frac{130^\circ-40^\circ}{2}=45^\circ}.$

12. A chord of a circle with a radius of 10 cm is 16 cm long. How far is it from the center?

Half of the chord is 8 cm, so the distance is ${\sqrt{10^2-8^2}=\sqrt{36}=6}$ cm.

13. A secant from a point $P$ meets a circle at $A$ and $B,$ with ${PA=3}$ cm and ${PB=12}$ cm. Find the length of the tangent from $P.$

${PT^2}={PA\cdot PB}={3\cdot 12}={36},$ so ${PT=6}$ cm.

More: Area formulas · Perimeter · Sphere · Triangles · Similar triangles