Theorems of circles

A circle is the set of all points that have fixed distance from a center. Circle theorems are about basic concepts of circles which include the rules of relevant angles, radius, chord, tangent and cyclic quadrilateral. In this article is the summary of 8 circle theorems.

Circle Theorem 1 - Inscribed angle theorem


If an inscribed angle and a central angle of a circle intercept the same arc, the inscribed angle is the half of the central angle.



To prove the theorem, consider 3 cases when the center of the circle is on one arm of the inscribed angle, between the arms of the inscribed angle and outside the inscribed angle.

Circle Theorem 2 - Angles in a semicircle


Measure of angles subtended to any point on the circumference of a circle from semicircle is 90 \degree .


The segment from the vertex of the angle to the center of the circle divides the triangle into two isosceles triangles .


Since the sum of all internal angles of a triangle is 180\degree ,


2\alpha +2\beta = 180\degree


Therefore,


\alpha +\beta = 90\degree


So the measure of the angle in the semicircle is 90\degree

Circle Theorem 3 - Angles in the same segment


The angles at the circumference subtended by the same arc are equal. Or Angles in the Same Segment are equal.


Angles at the circumference \alpha , \beta and the central angle \theta are subtended by the same arc.


According to the inscribed angles theorem,

\alpha = \dfrac{1}{2}\theta


\beta = \dfrac{1}{2}\theta


Therefore,

\alpha = \beta

Circle Theorem 4 - Cyclic quadrilateral


Opposite angles in a cyclic quadrilateral are supplementary which means their sum is equal to 180°


A cyclic quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle of the quadrilateral and the cyclic quadrilateral is also called inscribed quadrilateral of the circle.


\alpha =\dfrac{1}{2} \angle COB


\beta = \dfrac{1}{2} \angle BOC


Therefore,


\alpha + \beta = \dfrac{1}{2} (\angle COB +\angle BOC ) = 180 \degree


Similarly,


\delta + \gamma = 180 \degree

Circle Theorem 5 - Radius to a Tangent


The angle between a tangent to a circle and a radius from the center of the circle to the point of tangencye is 90°. The tangent is perpendicular to the radius drawn to the point of tangency.

Circle Theorem 6 - Two tangents from an external point to a circle

The lengths of the two tangents drawn from an external point to a circle are equal.


An external point A to the points of tangency B and C form two segment.


According to the theorem of radius of tangent, we know,


OB\perp AB , OC\perp AC


and


\triangle ABO and \triangle ACO are right triangles


Therefore,


AB = AC = \sqrt{AO^2-R^2}


In which R is the radius of the circle.

Circle Theorem 7 - Perpendicular to chord


The perpendicular from the center of a circle to a chord bisects the chord.

In the figure,


OC\perp AB

\because OA =OB

\therefore \triangle ABO is an isosceles triangle

\therefore AC = BC

Circle Theorem 8 - Alternate segment theorem


The angle between a tangent and a chord is equal to the angle in the alternate segment.


In the figure, the tangent AD and chord AB forms the angle \angle BAD

\angle BAD = \alpha


Join the center to point A and B . We have an isosceles triangle \triangle AOB


According to the Inscribed Angle Theorem, we know the measure of the inscribed angle is the half of the central angle.


So the alternative angle ACB has the following relationship with the central angle AOB


\angle ACB = \dfrac{1}{2} \angle AOB


According to the theorem of Radius to Tangent, \angle OAD = 90\degree


Therefore,


\angle OAB = OBA = 90-\alpha


Since all internal angles sum of a triangle is 180\degree ,


\angle AOB = 180\degree -2(90-\alpha) = 2\alpha


\therefore \angle ACB = \alpha

Steven Zheng Steven Zheng posted 3 months ago

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