In this article we have made a collection of 10 important circle theorems that you must know if you are a student or someone who wants to master everything about circle. Having read these theorems you will feel quite confident in solving the circle problems that you are often stuck on.
To understand these circle theorems you just need to have a basic idea of Circle fundamentals and triangle congruence and similarity. We first learn about the theorem followed by the proof for it. If you are not good at fundamentals of circle you can refer to another article on this portal that is basically dedicated to the Basics of Circle . So let's start with them one by one.
10 Circle theorems to get a hold on Geometry
To comprehend the nuances of mathematics we should focus on spending most of our time on concept building rather than starting with problem-solving.
Theorem 1 : Any line passing through the centre of the circle is always perpendicular to the tangent at the point of contact.
In the figure,
AB is a tangent and P is the point of contact.
OP is perpendicular to the tangent AB and line OP passing through the center.
Theorem 2 : Tangents drawn to the circle from a point outside the circle are always of equal in length.
Proof : we have to prove that AM = AN
Let's consider a point A outside the circle and draw two imaginary lines passing through the point A that touches the circle at two points M and N. O be the centre of the circle
Line AM and AN are tangents to the circle and we will prove they are equal.
Now consider the \(\Delta OMA \) and \(\Delta ONA \)
From theorem 1. AM is tangent to the circle hence \( \angle OMA \) and \( \angle ONA \) are \( 90°\)
\( \angle OMA\) =\( \angle ONA \) (Both 90°)
OA = OA ( common side and Hypotenuse in both the triangles )
OM = ON ( Radius of the circle)
By RHS congruence rule both the triangles
\(\Delta OMA \) and \(\Delta ONA \) are proven to be congruent.
\( \implies \) AM = AN
Theorem 3 : If a line passes through the centre and perpendicular to any chord on the circle it bisects the chord.
Proof : We have to prove AD = DB
Drop an imaginary perpendicular line from centre O to the chord AB. The perpendicular touches the chord at point D.
Now look at \(\Delta OAD \) and \(\Delta ODB \)
OA = OB ( Both are radius)
OD =OD ( common side in both triangles)
\( \angle ODA = \angle ODB \)
By RHS congruence rule \(\Delta OAD \) and \(\Delta ODB \) are congruent.
\(\therefore \) AD = DB (Both sides are equal)
Theorem 4 : Angle subtended by an arc at the centre is always double of the measure of the angle subtended by the same arc at any other point of the remaining part of the circle.
Proof : To Prove \( \angle ACB = {\angle AOB\over 2 }\)
In \(\Delta AOC \)
OC = OA (Both radii)
\( \therefore \) \(\Delta AOC \) is an isosceles triangle
Hence \( \angle ACO \) \(= \angle CAO \)
Outer \( \angle AOM \) \( = \angle ACO + \angle CAO \)
OC = OA (Both radii)
\( \therefore \) \(\Delta AOC \) is an isosceles triangle
Hence \( \angle ACO \) \(= \angle CAO \)
Outer \( \angle AOM \) \( = \angle ACO + \angle CAO \)
\( \implies \) \( \angle AOM = 2 \angle ACO \) ........Eq(1)
In \(\Delta BOC \)
In \(\Delta BOC \)
OC = OB (Both radii)
\( \therefore \) \(\Delta BOC \) is an isosceles triangle
Hence \( \angle BCO \) \(= \angle OBC \)
\( \therefore \) \(\Delta BOC \) is an isosceles triangle
Hence \( \angle BCO \) \(= \angle OBC \)
Outer \( \angle BOM \) \( = \angle BCO + \angle OBC \)
\( \implies \) \( \angle BOM = 2 \angle BCO \)..........Eq(2)
We can also write \( \angle AOB = \angle AOM + \angle BOM \)
From equation (1) and (2)
\( \angle AOB = 2 \angle ACO + 2 \angle BCO \)
From the figure \( \angle ACO + \angle BCO = \angle ACB \)
\( \angle AOB = 2 \angle ACB \) OR \( \angle ACB = {\angle AOB \over 2} \)
Theorem 5 : Angles formed on the same segment are always equal in measure.
Proof : Let AB is the chord and \(\Delta APB \) and \(\Delta AQB \) are the triangles formed on the same chord AB.
we will prove \( \angle APB \) = \( \angle AQB \)
From therorem 5 We can infer that the angle formed by segment AB at the center O is equal to twice the angle formed by the same segment at P.
\( \therefore \angle AOB = 2 \angle APB \) ..............Eq(1)
\( \therefore \angle AOB = 2 \angle APB \) ..............Eq(1)
From therorem 5 We can also conclude that the angle subtended by segment AB at the center O is equal to twice the angle subtended by the same segment at Q.
\( \therefore \angle AOB = 2 \angle AQB \) ..............Eq(2)
From Eq(1) and Eq(2)
\( 2 \angle APB = 2 \angle AQB \)
\( 2 \angle APB = 2 \angle AQB \)
\( \angle APB = \angle AQB \)
Theorem 6 : If two chords of a circle are equal in length they subtend equal angles at the centre of the circle.
Proof : To prove \( \angle AOB = \angle DOB \)
From the given figure
In \( \Delta AOB \) and \( \Delta DOC \)
OA = OC (Both radii)
OB = OD (Both radii)
Hence from SSS Congruent rule \( \Delta AOB \) and \( \Delta DOC \) are congruent .
\( \therefore \) \( \angle AOB = \angle DOB \)
Theorem 7 : If two chords subtend equal angles at the centre they are of equal length.
Proof : To prove AB =CD
Proof : To prove AB =CD
From the given figure
In \( \Delta AOB \) and \( \Delta DOC \)
OA = OC (Both radii)
OB = OD (Both radii)
Hence from SAS Congruent rule \( \Delta AOB \) and \( \Delta DOC \) are congruent .
\( \therefore \) AB = CD (Both chord are equal )
Theorem 8 : If two chords are equidistant from the centre, they are always equal in length.
Proof : To prove AB = CD
Proof : To prove AB = CD
From the figure given above
In right angle \( \Delta BOM \) and \( \Delta DON \)
OM = ON (both chords are equidistent from the centre )
\( \angle BMO = \angle DNO \) (90 degree \)
OB = OD (Both equal and both hypotenuse)
From RHS congruent rule both \( \Delta BOM \) and \( \Delta DON \) are congruent.
\( \therefore \) BM = DN
From Theorem 3 : we can write BM = 2AB and DN = 2CD
Hence 2AB = 2CD \( \implies \) AB = CD
Theorem 9 : If two chords are equal in length they are equidistant from the center.
Proof : To prove OM = ON
Proof : To prove OM = ON
From the figure look at the triangles BOM and DON
OB = OD
\( \angle BMO = \angle DON \) ( Both are 90 degree )
Given AB = CD
From theorem 3 : we can write BM = \( 1 \over 2 \) AB and DN = \(1 \over 2\) CD
\( \therefore \) BM = DN
From RHS congruent rule both \( \Delta BOM \) and \( \Delta DON \) are congruent.
\( \therefore \) OM = ON
Theorem 10 : A diameter of a circle subtends a right angle at the circumference of the circle.
Proof : \( \angle ABC = 90 \)
Proof : \( \angle ABC = 90 \)
consider triangle AOB
OB = OA (Both radii )
hence \( \Delta AOB \) is an isosceles triangle .
\(\angle OAB = \angle OBA \)
from angle sum property \( \angle AOB + \angle OAB + \angle OBA \) = 180
\( \angle AOB + 2 \angle OAB\) = 180
\( \angle AOB = 180 - 2 \angle OAB \) ....... Eq(1)
Now consider triangle BOC
OB = OC (Both radii )
hence \( \Delta BOC \) is an isosceles triangle .
\(\angle OBC = \angle OCB \)
from angle sum property \( \angle BOC + \angle OBC + \angle OCB \) = 180
\( \angle BOC + 2 \angle OCB\) = 180
\( \angle BOC = 180 - 2 \angle OCB \) ....... Eq(2)
we can also write
we can also write
\( \angle AOB + \angle BOC \) = 180
from equation (1) and equation (2)
\( 180 - 2 \angle OAB + 180 - 2 \angle OCB \) = 180
2 \( \angle OAB + \angle OCB \) = 180
2 \( \angle OAB + \angle OCB \) = 180
\( \angle OAB + \angle OCB \) = 90 OR \( \angle A + \angle C = 90 \) ........Eq(3)
Apply angle sum property in \( \Delta ABC \) also
\( \angle A + \angle B + \angle C \) = 180
From equation (3)
\( 90 + \angle B \) = 180
\( \angle B = 90 \) OR \( \angle ABC = 90 \)