Circle Theorems
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These are the solutions to the GCSE past questions on 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}
$
AQA GCSE Mathematics Higher Tier Formulae Sheet
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 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 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.
(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.
S is a point inside triangle PQR
Assume that S is the centre of the circle.
Work out the size of angle x
(b.) In fact, the centre of the circle is on PS but not at S.
What does this mean about the size of angle x?
Tick one box.
(c.) For a different circle,
AB is a tangent at A
C and D are on the circumference of the circle.
AC = CD
Here is Simon's method to work out the size of angle y.
Angle ADC = 70° (alternate segment theorem)
Therefore, y = 70° (angles in an isosceles triangle)
Is he correct?
Give a reason for your answer.
$ (a.) \\[3ex] \angle PSR = 2 \cdot \angle PQR ...\text{angle at center = twice angle at circumference} \\[3ex] 130 = 2 \cdot x \\[3ex] 2x = 130 \\[3ex] x = \dfrac{130}{2} \\[5ex] x = 65^\circ \\[3ex] $ (b.) The angle on a straight line is 180°
If the center of the circle is on a straight line (180°), then the value of x = 180 ÷ 2 = 90°
90° is > 65°
Hence, it is greater than the answer to part (a.)
(c.) He is not correct.
This is because the alternate segment theorem applies to Angle ACD (not Angle ADC)
A, B and C are points on the circumference of a circle, centre O
AOB is a diameter of the circle.
Prove that angle ACB is 90°
You must not use any circle theorems in your proof.
Using Triangle Theorems:
Construction: Draw the radius from the centre O to point C on the circumference
This will display two triangles: $\triangle AOC$ and $\triangle BOC$
Label the angles in the triangle.
$ \underline{\triangle AOC} \\[3ex] Because: \\[3ex] |AO| = |CO| ...same\;\;radius \\[3ex] \angle OAC = \angle OCA = \theta ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOC \\[3ex] \angle OAC + \angle OCA + \angle AOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle AOC \\[3ex] \implies \\[3ex] \theta + \theta + \alpha = 180 \\[3ex] 2\theta + \alpha = 180 ...eqn.(1) \\[5ex] \underline{\triangle BOC} \\[3ex] Because: \\[3ex] |BO| = |CO| ...same\;\;radius \\[3ex] \angle OBC = \angle OCB = \phi ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOC \\[3ex] \angle OBC + \angle OCB + \angle BOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle BOC \\[3ex] \implies \\[3ex] \phi + \phi + \beta = 180 \\[3ex] 2\phi + \beta = 180 ...eqn.(2) \\[5ex] \alpha + \beta = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \alpha + \beta = 180...eqn.(3) \\[5ex] eqn.(1) + eqn.(2) \implies \\[3ex] (2\theta + \alpha) + (2\phi + \beta) = 180 + 180 \\[3ex] 2\theta + \alpha + 2\phi + \beta = 360 \\[3ex] 2\theta + 2\phi + \alpha + \beta = 360 \\[3ex] Substitute\;\;for\;\;eqn.(3) \\[3ex] 2\theta + 2\phi + 180 = 360 \\[3ex] 2\theta + 2\phi = 360 - 180 \\[3ex] 2\theta + 2\phi = 180 \\[3ex] 2(\theta + \phi) = 180 \\[3ex] \theta + \phi = \dfrac{180}{2} \\[5ex] \theta + \phi = 90^\circ \\[3ex] \angle ACB = \theta + \phi ...diagram \\[3ex] \therefore \angle ACB = 90^\circ $
E, F and G are points on a circle with centre H
$E\hat{F}H = x$ and $G\hat{F}H = y$
Complete the proof of the following statement:
Tha angle at the centre is twice the angle at the circumference.
Proof:
$E\hat{F}G = x + y$
$F\hat{E}H = x$ (triangle FEH is isosceles)
$F\hat{H}E = .................................$
$ (b.) \\[3ex] For\;\;this\;\;question: \\[3ex] \underline{Given}: \\[3ex] E\hat{F}G = x + y ...\angle \;\;at\;\;circumference \\[3ex] Obtuse\;\; \angle EHG = \angle \;\;at\;\;centre \\[3ex] \underline{To\;\;Prove}: \\[3ex] \angle EHG = 2(x + y) \\[3ex] \underline{\triangle FEH} \\[3ex] E\hat{F}H = x...as\;\;shown\;\;in\;\;the\;\;diagram \\[3ex] F\hat{E}H = E\hat{F}H = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle FEH \\[3ex] F\hat{H}E + F\hat{E}H + H\hat{F}E = 180...sum\;\;of\;\;\angle s\;\;of\;\;\triangle FEH \\[3ex] F\hat{H}E + x + x = 180 \\[3ex] F\hat{H}E + 2x = 180 \\[3ex] F\hat{H}E = 180 - 2x ...eqn.(1) \\[3ex] \underline{\triangle FEG} \\[3ex] H\hat{F}G = y...as\;\;shown\;\;in\;\;the\;\;diagram \\[3ex] F\hat{G}H = H\hat{F}G = y ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle FEH \\[3ex] F\hat{H}G + F\hat{G}H + H\hat{F}G = 180...sum\;\;of\;\;\angle s\;\;of\;\;\triangle FEH \\[3ex] F\hat{H}G + y + y = 180 \\[3ex] F\hat{H}G + 2y = 180 \\[3ex] F\hat{H}G = 180 - 2y ...eqn.(2) \\[3ex] Reflex\;\; \angle EHG = F\hat{H}E + F\hat{H}G \\[3ex] Reflex\;\; \angle EHG = (180 - 2x) + (180 - 2y) \\[3ex] Reflex\;\; \angle EHG = 180 - 2x + 180 - 2y \\[3ex] Reflex\;\; \angle EHG = 360 - 2x - 2y \\[3ex] Reflex\;\; \angle EHG + Obtuse\;\; \angle EHG = 360^\circ...sum\;\;of\;\;\angle s\;\;at\;\;a\;\;point \\[3ex] (360 - 2x - 2y) + Obtuse\;\; \angle EHG = 360 \\[3ex] Obtuse\;\;\angle EHG = 360 - (360 - 2x - 2y) \\[3ex] Obtuse\;\;\angle EHG = 360 - 360 + 2x + 2y \\[3ex] Obtuse\;\;\angle EHG = 2x + 2y \\[3ex] Obtuse\;\;\angle EHG = 2(x + y) \\[3ex] \implies \\[3ex] $ The angle at the centre is twice the angle at the circumference.
