Circle Theorems
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Length of Arc, Area of Sector, Area of Circle, Circumference of Circle
Except stated otherwise, use:
$
d = diameter \\[3ex]
r = radius \\[3ex]
L = arc\:\:length \\[3ex]
A = area\;\;of\;\;sector \\[3ex]
\theta = central\:\:angle \\[3ex]
\pi = \dfrac{22}{7} \\[5ex]
RAD = radians \\[3ex]
^\circ = DEG = degrees \\[7ex]
\underline{\theta\;\;in\;\;DEG} \\[3ex]
L = \dfrac{2\pi r\theta}{360} \\[5ex]
\theta = \dfrac{180L}{\pi r} \\[5ex]
r = \dfrac{180L}{\pi \theta} \\[5ex]
A = \dfrac{\pi r^2\theta}{360} \\[5ex]
\theta = \dfrac{360A}{\pi r^2} \\[5ex]
r = \dfrac{360A}{\pi\theta} \\[7ex]
\underline{\theta\;\;in\;\;RAD} \\[3ex]
L = r\theta \\[5ex]
\theta = \dfrac{L}{r} \\[5ex]
r = \dfrac{L}{\theta} \\[5ex]
A = \dfrac{r^2\theta}{2} \\[5ex]
\theta = \dfrac{2A}{r^2} \\[5ex]
r = \sqrt{\dfrac{2A}{\theta}} \\[7ex]
Circumference\:\:of\:\:a\:\:circle = 2\pi r = \pi d \\[3ex]
L = \dfrac{2A}{r} \\[5ex]
r = \dfrac{2A}{L} \\[5ex]
A = \dfrac{Lr}{2}
$
Circle Theorems
(1.) The angle in a semicircle is a right angle (an angle of 90°).
(2.) Angles in the same segment of a circle are equal.
OR
Angles subtended by a chord of a circle in the same segment of the circle are equal.
(3.) The angle which an arc of a circle subtends at the center is twice the angle which the same
arc of the circle subtends at the circumference.
OR
The measure of any angle inscribed in a circle is half the measure of the intercepted arc.
(4.) The sum of the interior opposite angles of a cyclic quadrilateral is 180°
OR
The interior opposite angles of a cyclic quadrilateral are supplementary
(5.) The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
(6.) The radius of a circle is perpendicular to the tangent of the circle at the point of contact.
This implies that the angle between the radius of a circle and the tangent to the circle at the point of
contact is 90°
(7.) Intersecting Tangents Theorem or Intersecting Tangent-Tangent Theorem and
Angle of Intersecting Tangents Theorem
If two tangents are drawn from the same external point:
(a.) the two tangents are equal in length
(b.) the line joining the external point and the centre of the circle bisects the angle formed by the two
tangents.
(c.) the line joining the external point and the centre of the circle bisects the angle formed by the two
radii.
(8.) Alternate Segment Theorem
The angle between a tangent to a circle and a chord drawn from the point of contact, is equal to the angle in
the alternate segment.
(9.) If a line drawn from the center of the circle bisects a chord, then:
(a.) it bisects its arc (the angle opposite the chord) and
(b.) it is perpendicular to the chord.
(10.) If a line drawn from the center of the circle is perpendicular to a chord, then:
(a.) it bisects the chord and
(b.) it bisects its arc (the angle opposite the chord).
(11.) Intersecting Chords Theorem
When two chords intersect, the product of the lengths of the segments of one chord is equal to the product of
the lengths of the segments of the other chord.
(12.) Angle of Intersecting Chords Theorem
The angle formed when two chords intersect is equal to half the sum of the intercepted arcs.
(13.) Intersecting Secants Theorem or Intersecting Secant-Secant Theorem
In the intersection of two secants from the same exterior point:
the product: of the distance between the first point and the external point and the distance between
the second point and the external point for the first secant is equal to
the product of the distance between the first point and the external point and the distance between the
second point and the external point for the second secant.
Alternatively, we can state it as: If two secant segments are drawn to a circle from an exterior point, then
the product of the measures of one secant segment and its external segment is equal to the product of the
measures of the other secant segment and its external segment.
(14.) Angle of Intersecting Secants (Inside the Circle) Theorem
The angle formed when two secants intersect inside a circle is equal to half the sum of the intercepted arcs.
(15.) Angle of Intersecting Secants (Outside the Circle) Theorem
The angle formed when two secants intersect outside a circle is equal to half the difference of the
intercepted arcs.
(16.) Intersecting Secant-Tangent Theorem or Intersecting Tangent-Secant Theorem
In the intersection of a secant and a tangent from the same external point:
the product of the distance between the first point and the external point and the distance between the
second point and the external point for the secant is equal to
the square of the distance between the point of contant and the external point for the tangent.
(17.) Angle of Intersecting Secant-Tangent Theorem
The angle formed when a secant and a tangent intersect outside a circle is equal to half the difference of the
intercepted arcs.
PC = 10.
Let $m\angle CBD = (x)^\circ$ and $m\angle BCD = (x + 54)^\circ$.
(A.) Find the value of x
(B.) The length of $\overline{CD}$ is .......... because ..........
1st option: 10; less than 10; greater than 10
2nd option: $\triangle CPD$ is equilateral; $m\angle CPD \lt 60^\circ$; $m\angle CPD \gt 60^\circ$
$ (A.) \\[3ex] m\angle BCD = 90^\circ ...\text{angle in a semicircle} \\[3ex] m\angle BCD = (x + 54)^\circ ...\text{Given} \\[3ex] \implies \\[3ex] x + 54 = 90 \\[3ex] x = 90 - 54 \\[3ex] x = 36^\circ \\[5ex] (B.) \\[3ex] |PB| = |PC| = |PD| = 10\;units...\text{radii} \\[3ex] \underline{\triangle PCD} \\[3ex] \triangle PCD \text{ is isosceles because } |PC| = |PD|...\text{radii} \\[3ex] |CD| \lt (|PC| + |PD|) ...\text{Triangle Inequality Theorem} \\[3ex] |CD| \lt (10 + 10) \\[3ex] |CD| \lt 20 \\[3ex] \text{1st Option:} \\[3ex] 20 \gt 10 \\[3ex] \implies \\[3ex] |CD| \gt 10 \\[3ex] $ Alternative Explanation
|CD| cannot be less than 10 because that would invalidate the Triangle Inequality Theorem
|CD| is not equal to 10 bbecause there is no information that suggests that $\triangle PCD$ is equilateral
So, |CD| is greater than 10
$ \text{2nd Option:} \\[3ex] m\angle CBD = m\angle CBP = x^\circ ...\text{diagram} \\[3ex] m\angle CBP = m\angle BCP = x^\circ ...\text{base angles of isosceles triangle PCB} \\[5ex] m\angle BCD = m\angle BCP + m\angle PCD ...\text{diagram} \\[3ex] x + 54 = x + m\angle PCD \\[3ex] m\angle PCD = x + 54 - x \\[3ex] m\angle PCD = 54^\circ \\[5ex] \underline{\triangle PCD} \\[3ex] m\angle PDC = m\angle PCD = 54^\circ ...\text{base angles of isosceles triangle PCD} \\[3ex] m\angle PDC + m\angle PCD + m\angle CPD = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] 54 + 54 + m\angle CPD = 180 \\[3ex] m\angle = 180 - 54 - 54 \\[3ex] m\angle CPD = 72^\circ \\[3ex] m\angle \gt 60^\circ \\[3ex] $ Alternative Explanation
Side Length - Angle Measure Theorem:
If any two side lengths of a triangle are unequal; the angles of the triangle are also unequal, and the measure of an angle is opposite the length of the side facing that angle as regards size.
The measure of the smallest angle is opposite the shortest side length.
The measure of the greatest angle is opposite the longest side length.
The measure of the middle angle is opposite the middle side length.
In other words; regarding size, the measure of an angle is opposite the length of the side facing the angle; or the side length facing an angle is opposite the angle measure as regards size.
Small side faces small angle, middle side faces middle angle, big side faces big angle
Because |CD| is greater than 10, the angle opposite |CD| should be the greatest angle.
60° is the angle of an equilateral triangle
For an isosceles triangle, the congruent angle is less than 60°
For an isosceles triangle, the sum of the congruent angles is less than 120°
180° − 120° = 60° (but note that it is not equal to 120°)
This implies that the 3rd angle is greater than 60°
This implies that $m\angle CPD$ is greater than 60°
AC is perpendicular to BD and $\angle CAD = \theta$.
Determine $\angle ABC$
A Cyclic Quadrilateral is a quadrilateral whose four vertices lie on the circumference of the circle.
It is also known as an Inscribed Quadrilateral
Let us represent the information diagrammatically
$ \underline{\triangle CAD} \\[3ex] \angle ACD = \angle CAD = \theta ...\text{base angles of isosceles triangle CAD} \\[3ex] \angle ACD + \angle CAD + \angle ADC = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] \theta + \theta + \angle ADC = 180 \\[3ex] \angle ADC = 180 - \theta - \theta \\[3ex] \angle ADC = 180 - 2\theta \\[5ex] \underline{\text{Cyclic Quadrilateral ABCD}} \\[3ex] \angle ABC + \angle ADC = 180^\circ ...\text{interior opposite angles of a cyclic are supplementary} \\[3ex] \angle ABC = 180 - \angle ADC \\[3ex] \angle ABC = 180 - (180 - 2\theta) \\[3ex] \angle ABC = 180 - 180 + 2\theta \\[3ex] \angle ABC = 2\theta $
(a.) Find $m\angle M$
(b.) Find $m\overset{\huge\frown}{MJ}$
Let us make some constructions in the diagram that will assist us to use circle theorems to solve the question
$ \angle CJK = \angle CLK = 90^\circ ...\text{radius is } \perp \text{ to tangent at point of contact} \\[3ex] |JK| = |LK| ...\text{Intersecting Tangents Theorem} \\[3ex] \angle JKC = \angle LKC = \dfrac{40^\circ}{2} = 20^\circ ...\text{Angle of Intersecting Tangents} \\[5ex] \underline{\triangle JCK} \\[3ex] \angle JCK + \angle CJK + \angle JKC = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] \angle JCK + 90 + 20 = 180 \\[3ex] \angle JCK = 180 - 110 \\[3ex] \angle JCK = 70^\circ \\[5ex] \angle LCK = \angle JCK = 70^\circ ...\text{Intersecting Tangents Theorem} \\[3ex] \angle JCL = \angle JCK + \angle LCK ...\text{Diagram} \\[3ex] \angle JCL = 70 + 70 \\[3ex] \angle JCL = 140^\circ \\[5ex] (a.) \\[3ex] \angle JCL = 2 * \angle M ...\text{angle at center = twice angle at circumference} \\[3ex] 140 = 2 * \angle M \\[3ex] \angle M = \dfrac{140}{2} \\[5ex] m\angle M = 70^\circ \\[5ex] (b.) \\[3ex] Reflex\;\angle JCL + Obtuse\;\angle JCL = 360^\circ ...\text{angles at a point} \\[3ex] Reflex\;\angle JCL + 140 = 360 \\[3ex] Reflex\;\angle JCL = 360 - 140 \\[3ex] Reflex\;\angle JCL = 220^\circ \\[5ex] \overset{\huge\frown}{ML} + \overset{\huge\frown}{MJ} = Reflex\;\angle JCL ...\text{intercepted arc = angle at center} \\[3ex] 138 + \overset{\huge\frown}{MJ} = 220 \\[3ex] \overset{\huge\frown}{MJ} = 220 - 138 \\[3ex] m\overset{\huge\frown}{MJ} = 82^\circ $
GBHD, OAG and OHC are straight lines.
$ HD = x\;cm \\[3ex] HB = 14\;cm \\[3ex] GB = 10\;cm \\[3ex] GA = 8\;cm \\[3ex] HC = 4\;cm \\[3ex] \text{Find the value of } x \\[3ex] $
Let us do some constructions to solve the question.
Let radius of the circle = r
$ |OK| = |OC| = |OA| = |OP| = r ...\text{radius of the circle} \\[3ex] |OH| + |HC| = |OC| ...\text{diagram} \\[3ex] |OH| = |OC| - |HC| \\[3ex] |OH| = r - 4 \\[5ex] \underline{\text{Intersecting Secants Theorem}} \\[3ex] \text{Exterior Point} = G \\[3ex] \text{1st Secant}: |GD| \\[3ex] |GD| = |GB| + |BH| + |HD| \\[3ex] |GD| = 10 + 14 + x \\[3ex] |GD| = 24 + x \\[3ex] \underline{\text{1st Product}} \\[3ex] |GB| * |GD| \\[3ex] |GB| * |GD| = 10(24 + x) \\[5ex] \text{2nd Secant}: |GP| \\[3ex] |GP| = |GA| + |AO| + |OP| \\[3ex] |GP| = 8 + r + r \\[3ex] |GP| = 8 + 2r \\[3ex] \underline{\text{2nd Product}} \\[3ex] |GA| * |GP| \\[3ex] |GA| * |GP| = 8(8 + 2r) \\[5ex] \text{Equate the two products} \\[3ex] 10(24 + x) = 8(9 + 2r) \\[3ex] 5(24 + x) = 4(8 + 2r) \\[3ex] 120 + 5x = 32 + 8r \\[3ex] 8r = 120 + 5x - 32 \\[3ex] r = \dfrac{5x + 88}{8} ...eqn(1) \\[5ex] \underline{\text{Intersecting Chords Theorem}} \\[3ex] \text{Chords |BD| and |KC|} \\[3ex] 14 * x = 4[r + (r - 4)] \\[3ex] 14x = 4(r + r - 4) \\[3ex] 14x = 4(2r - 4) \\[3ex] 14x = 8r - 16 \\[3ex] 8r - 16 = 14x \\[3ex] 8r = 14x + 16 \\[3ex] r = \dfrac{14x + 16}{8} ...eqn.(2) \\[5ex] r = r \implies eqn.(1) = eqn.(2) \\[3ex] \dfrac{5x + 88}{8} = \dfrac{14x + 16}{8} \\[5ex] 5x + 88 = 14x + 16 \\[3ex] 14x - 5x = 88 - 16 \\[3ex] 9x = 72 \\[3ex] x = \dfrac{72}{9} \\[5ex] x = 8\;cm $
Find the value of x
Let us indicate points on the circle
$ \underline{\triangle AOD} \\[3ex] \angle OAD = \angle ODA = p ...\text{base angles of isosceles triangle} \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] p + p + 5x = 180 \\[3ex] 2p = 180 - 5x \\[3ex] p = \dfrac{180 - 5x}{2} \\[5ex] \implies \\[3ex] \angle OAD = \angle ODA = \dfrac{180 - 5x}{2} \\[5ex] \underline{\text{Cyclic Quadrilateral ABCD}} \\[3ex] \angle BAD + \angle BCD = 180^\circ ...\text{interior opposite angles of a cyclic quad are supplementary} \\[3ex] \angle BAD + 6x = 180 \\[3ex] \angle BAD = 180 - 6x...eqn.(1) \\[5ex] \angle AOD = 2 * \angle ABD ...\text{angle at center = twice the angle at circumference} \\[3ex] 5x = 2 * \angle ABD \\[3ex] \angle ABD = \dfrac{5x}{2}...eqn.(2) \\[5ex] \angle BDA = \angle BDO + \angle ODA ...\text{diagram} \\[3ex] \angle BDA = x + \dfrac{180 - 5x}{2}...eqn.(3) \\[5ex] \underline{\triangle ABD} \\[3ex] eqn.(1) + eqn.(2) + eqn.(3) = 180 \\[3ex] \angle BAD + \angle ABD + \angle BDA = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] 180 - 6x + \dfrac{5x}{2} + x + \dfrac{180 - 5x}{2} = 180 \\[5ex] -5x + \dfrac{5x}{2} + \dfrac{180}{2} - \dfrac{5x}{2} = 0 \\[5ex] -5x + 90 = 0 \\[3ex] 90 = 5x \\[3ex] 5x = 90 \\[3ex] x = \dfrac{90}{5} \\[5ex] x = 18^\circ $
Use the diagram to answer the questions.
(a.) If $m\angle BOF = 50^\circ$, find $m\overset{\huge\frown}{BF}$
(b.) If $m\angle BOC = 70^\circ$, find $m\angle C$
(c.) If $BE = 26$ and $DF = 24$, find $OX$
(d.) If $m\overset{\huge\frown}{DEF} = 250^\circ$, find $m\overset{\huge\frown}{BD}$
(e.) If $m\overset{\huge\frown}{BF} = 80^\circ$, find $m\overset{\huge\frown}{BEF}$
(f.) If $m\angle FO = 140^\circ$, find $m\angle FDE$
$ (a.) \\[3ex] m\overset{\huge\frown}{BF} = \angle BOF = 50^\circ \\[3ex] ...\text{intercepted arc = central angle} \\[5ex] (b.) \\[3ex] \angle OBC = 90^\circ \\[3ex] ...\text{radius |OB| is } \perp \text{ to tangent |ABC| at point of contact: B} \\[3ex] \underline{\triangle OBC} \\[3ex] \angle BCO + \angle OBC + \angle BOC = 180^\circ \\[3ex] ...\text{sum of interior angles in a triangle} \\[3ex] \angle BCO = \angle C ...\text{diagram} \\[3ex] \angle C + 90 + 70 = 180 \\[3ex] \angle C = 180 - 90 - 70 \\[3ex] \angle C = 20^\circ \\[5ex] (c.) \\[3ex] |BE| = 26...\text{diameter} \\[3ex] |BO| = |OE| = \dfrac{|BE|}{2} = \dfrac{26}{2} = 13...\text{radii} \\[5ex] |OE| = |OF| = |OD| = |OB| = 13...\text{radii} \\[3ex] |XF| = |XD| = \dfrac{|DF|}{2} = \dfrac{24}{2} = 12 \\[5ex] ...\perp\text{ bisector |BE| of chord |DF| passes through the center} \\[3ex] \underline{\text{Right Triangle FOX}} \\[3ex] |OF|^2 = |OX|^2 + |XF|^2 ...\text{Pythagorean Theorem} \\[3ex] |OX|^2 = |OF|^2 - |XF|^2 \\[3ex] |OX|^2 = 13^2 - 12^2 \\[3ex] |OX| = \sqrt{169 - 144} \\[3ex] |OX| = 5 \\[5ex] (d.) \\[3ex] \text{Reflex }\angle FOD = m\overset{\huge\frown}{DEF} = 250^\circ \\[3ex] ...\text{central angle = intercepted arc} \\[5ex] \text{Obtuse }\angle FOD + \text{Reflex }\angle FOD = 360^\circ ...\text{Angles Around a Point Theorem} \\[3ex] \text{Obtuse }\angle FOD = 360 - \text{Reflex }\angle FOD \\[3ex] = 360 - 250 \\[3ex] = 110^\circ \\[5ex] \text{chord} = |DF| \\[3ex] \text{central angle subtended by chord} = \angle FOD = 110^\circ \\[3ex] \perp \text{bisector of chord |DF|} = |BE| \\[3ex] \perp \text{bisector of chord that passes through the center bisects the central angle subtended by chord} \\[3ex] \angle XOF = \angle XOD = \dfrac{\angle FOD}{2} = \dfrac{110}{2} = 55^\circ \\[5ex] m\overset{\huge\frown}{BD} = \angle BOD = 55^\circ \\[3ex] ...\text{intercepted arc = central angle} \\[5ex] (e.) \\[3ex] \angle \overset{\huge\frown}{BF} = 2 * \angle BEF \\[3ex] ...\text{intercepted arc = twice the inscribed angle} \\[3ex] 2 * \angle BEF = 80 \\[3ex] \angle BEF = \dfrac{80}{2} = 40^\circ \\[5ex] (f.) \\[3ex] \angle FOE = 2 * \angle FDE \\[3ex] \text{central angle = twice the inscribed angle} \\[3ex] 2 * \angle FDE = 140 \\[3ex] \angle FDE = \dfrac{140}{2} \\[5ex] \angle FDE = 70^\circ $
