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Circle Geometry, SMB-019

The diagram shows a large semicircle with diameter  `AB`  and two smaller semicircles with diameters  `AC`  and  `BC`, respectively, where  `C`  is a point on the diameter  `AB`. The point  `M`  is the centre of the semicircle with diameter  `AC`.

The line perpendicular to  `AB`  through  `C`  meets the largest semicircle at the point  `D`. The points  `S`  and  `T`  are the intersections of the lines  `AD`  and  `BD`  with the smaller semicircles. The point  `X`  is the intersection of the lines  `CD`  and  `ST`.
 

Explain why `CTDS` is a rectangle.   (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`/_SDT = 90°\ text{(angle in semi-circle)}`

`/_ ASC = 90°\ \ text{(angle in semi-circle)}`

`=> /_ CSD = 180-90=90°\ \ text{(} /_ ASD\ text{is a straight line)}`

`text(Similarly,)`

`/_CTB = /_CTD=90°`

`/_SCT = 90°\ \ text{(angle sum of quadrilateral}\ CTDS text{)}`

`text(S)text(ince all angles are right angles,)`

`CTDS\ text(is a rectangle)` 

Circle Geometry, SMB-018

In the diagram, `AB` is a diameter of a circle with centre `O`. The point `C` is chosen such that `Delta ABC` is acute-angled. The circle intersects `AC` and `BC` at `P` and `Q` respectively.
 

Why is `/_BAC = /_CQP`?   (2 marks)

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`text(Proof)\ text{(See Worked Solutions)}`

 

Show Worked Solution

`/_BAC + /_BQP = 180°\ \ (APQB\ text{is a cyclic quad})`

`/_CQP + /_BQP = 180°\ text{(}/_ CQB\ text{is a straight line)}`

`:.\ /_BAC = /_CQP`

Circle Geometry, SMB-017

In the diagram, the points `A`, `B`, `C` and `D` are on the circumference of a circle, whose centre `O` lies on `BD`. The chord `AC` intersects the diameter `BD` at `Y`. The tangent at `D` passes through the point `X`.

It is given that  `∠CYB = 100^@`  and  `∠DCY = 30^@`.

 

 

  1. What is the size of  `∠ACB`?   (1 mark)

  2. What is the size of  `∠CBD`?   (2 marks)

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  1. `60^@`
  2. `20^@`
Show Worked Solution
i.    `∠DCB` `= 90^@\ \ text{(angle in semi-circle)}`
`∠ACB` `= 90-30`
  `= 60^@`

 
ii.    `∠CYD = 180-100=80^@\ \ text{(180° in straight line)}`

`∠CDY = 180-(80+30)=70^@\ \ text{(180° in Δ)}`

`∠CBD = 180-(90+70)=20^@\ \ text{(180° in Δ)}`

Circle Geometry, SMB-016

The line \(BD\) is a tangent to a circle and the secant \(AD\) intersects the circle at \(A\) and \(C\).
 

Given that  \(AC = 18\)  and  \(CD = 6\), find the value of \(x\).  (2 marks)   

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\(x=12\)

Show Worked Solution
\(x^2\) \(= 6 \times (18 + 6) \)  
  \(=144\)  
\(x\) \(=12\)  

Circle Geometry, SMB-015

 

Find the size of angles \(x^{\circ}\) and \(y^{\circ}\).  (3 marks)   

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\(x= 38^{\circ}\)

\(y= 104^{\circ}\)

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\(x=38^{\circ}\ \ \text{(angles standing on the same arc)} \)

\(\text{Let}\ \ \theta =\ \text{angle at centre on the same arc} \)

\(\theta = 2 \times 38 = 76^{\circ} \)

\(y^{\circ}\) \(=180-76\ \ \text{(180° in straight line)}\)  
  \(=104^{\circ} \)  

Circle Geometry, SMB-014

Find \(\angle ADC\).  (2 marks)   

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\(x= 84^{\circ}\)

Show Worked Solution

\(\angle ABC = 180-105=75^{\circ}\ \ \text{(straight line)} \)

\(\angle ADC\) \(=180-75\ \ \text{(opposite angles of cyclic quad are supplementary)}\)  
  \(=105^{\circ} \)  

Circle Geometry, SMB-012

In the diagram, \(AC\) is a diameter of the circle centred \(O\), \(\angle BAC = 20^{\circ}\) and \(\angle CAD = 62^{\circ} \).
 

Find \(\angle BCD\).  (2 marks)   

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\(\angle BCD= 98^{\circ}\)

Show Worked Solution

\( \angle BCD + \angle DAB = 180^{\circ} \ \ \text{(opposite angles of cyclic quad)} \)

\(\angle BCD\) \(=180-(62+20)\)  
  \(=98^{\circ} \)  

Circle Geometry, SMB-011

In the diagram, \(PR\) is a diameter of the circle centred \(O\) and \(\angle QPR = 15^{\circ} \).
 

Find \(\theta\).  (2 marks)   

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\(\theta = 75^{\circ}\)

Show Worked Solution

\(\angle PQR = 90^{\circ}\ \  \text{(angle in semicircle)} \)

\(\theta\) \(=180-(90+15)\ \ (180^{\circ}\ \text{in}\ \Delta) \)  
  \(=75^{\circ} \)  

Circle Geometry, SMB-010

In the circle with centre at \(O\), \(\angle BAC = 36^{\circ}\).
 

Find \(\theta\).  (2 marks)   

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\(\theta = 72^{\circ}\)

Show Worked Solution

\(\text{Property: angle at centre is twice angle on circumference, standing on same arc.}\)

\(\theta= 2 \times 36= 72^{\circ}\)

Circle Geometry, SMB-009

In the diagram, \(OB\) meets the chord \(AC\) such that \(AB = BC\).

The length of chord \(AC = 24\), and \(OC = 13\). 
 

Find the length of \(OB\).  (3 marks)   

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\(OB=5\)

Show Worked Solution

\(OB \perp AC\ \ \text{(line through centre that bisects chord)}\)

\(BC= \dfrac{1}{2} \times 24 = 12 \)

\(\text{Using Pythagoras in}\ \Delta OBC :\)

\(OB^2\) \(= 13^2-12^2 \)  
  \(= 25\)  
\(\therefore OB\) \(=5\)  

Circle Geometry, SMB-008

In the diagram, \(AC\) is a diameter of the circle centred at \(O\), and \(OA = AB\).
 

Find the value of \(\theta\).   (3 marks)   

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\(\theta = 30^{\circ}\)

Show Worked Solution

\(\angle ABC=90^{\circ}\ \ \text{(angle in semi-circle)}\)

\(OA=OB\ \ \text{(radii)} \)

\( \angle OAB=60^{\circ}\ \ ( \Delta OAB\ \text{is equilateral}) \)

\(\theta\) \(= 180-(90+60)\ \ (180^{\circ}\ \text{in}\ \Delta) \)  
  \(= 30^{\circ} \)  

Cicle Geometry, SMB-007

In the diagram, a line from the centre of the circle meets a chord at its midpoint.

Find the value of \(\theta\).  (2 marks) 
  

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\(\theta = 47^{\circ}\)

Show Worked Solution

\(\text{Line from centre bisects chord}\ \ \Rightarrow\ \ \text{Line is ⊥ to chord}\)

\(\theta\) \(= 180-(90+43)\ \ (180^{\circ}\ \text{in}\ \Delta) \)  
  \(= 47^{\circ} \)  

Circle Geometry, SMB-006

In the circle centred at \(O\), the chord \(AC\) has length 15 and \(OB\) meets the chord \(AC\) at right angles.

Find the length of \(BC\).  (1 mark) 
  

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\(BC = 7.5\)

Show Worked Solution
\(BC\) \(= \dfrac{1}{2} \times 15\ \ \text{(perpendicular from centre to chord bisects chord)}\)  
  \(= 7.5 \)  

Circle Geometry, SMB-003

In the diagram, the vertices of  \(\Delta ABC\)  lie on the circle with centre \(O\). The point \(D\) lies on \(BC\) such that \(\Delta ABD\) is isosceles and \(\angle ABC = x\).

Explain why \(\angle AOC =2x\).   (2 marks)

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\(\text{See Worked Solutions}\)

Show Worked Solution

\(\text{Angles at circumference and centre are both on arc}\ AC\)

\(\therefore \angle AOC = 2x\)

Circle Geometry, SMB-002

The line \(AT\) is the tangent to the circle at \(A\), and \(BT\) is a secant meeting the circle at \(B\) and \(C\).
  

Given that  \(AT = 12\),  \(BC = 7\)  and  \(CT = x\), find the value of \(x\).  (2 marks)

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\(x = 9\)

Show Worked Solution

\(\text{Property: square of tangent = product of secant intercepts}\)

\(AT^2\) \(= CT \times BT\)
\(12^2\) \(= x(x + 7)\)
\(144\) \(= x^2 + 7x\)
\(x^2 + 7x-144\) \(= 0\)
\((x + 16)(x-9)\) \(= 0\)

 

\(\therefore x = 9,\  (x \gt 0) \)

Circle Geometry, SMB-001

In the circle centred at \(O\) the chord \(AB\) has length 7. The point \(E\) lies on \(AB\) and \(AE\) has length 4. The chord \(CD\) passes through \(E\).

Let the length of \(CD\) be \(\ell\) and the length of \(DE\) be \(x\).

Show that  \(x^2-\ell x + 12 = 0\).  (2 marks)

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\(\text{See Worked Solutions}\)

Show Worked Solution

\(\text{Show}\ \ x^2-\ell x + 12 = 0\)
 


 

\(AB = 7 \ \ \Rightarrow \ EB=7-4=3\)

\(AE \times EB = ED \times CE\ \ \text{(intercepts of intersecting chords)}\)

\(4 \times 3\) \(= x(\ell-x)\)
\(12\) \(= x\ell-x^2\)
\(\therefore x^2-\ell x+12\) \(=0\)

Plane Geometry, EXT1 2018 HSC 11d

Two secants from the point `C` intersect a circle as shown in the diagram.
 

 
What is the value of `x`?  (2 marks)

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`4`

Show Worked Solution

`text(Using formula for intercepts of intersecting secants:)`

`x (x + 2)` `= 3 (3 + 5)`
`x^2 + 2x` `= 24`
`x^2 + 2x – 24` `= 0`
`(x + 6) (x – 4)` `= 0`
`:. x` `= 4 \ \ \ (x > 0)`

Plane Geometry, EXT1 2017 HSC 12a

The points `A`, `B` and `C` lie on a circle with centre `O`, as shown in the diagram. The size of `angleAOC` is 100°.
 

Find the size of `angleABC`, giving reasons.  (2 marks)

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`130^@`

Show Worked Solution

`text(Reflex)\ angleAOC` `= 360-100`
  `= 260^@`
`:. angleABC` `= 1/2 xx 260^@` `text{(angles at centre and on}`
 `text{circumference of arc}\ AC)`
  `= 130^@`  

Plane Geometry, EXT1 2016 HSC 4 MC

In the diagram, `O` is the centre of the circle `ABC`, `D` is the midpoint of `BC`, `AT` is the tangent at `A` and  `∠ATB = 40^@`.
 

What is the size of the reflex angle `DOA`?

  1. `80^@`
  2. `140^@`
  3. `220^@`
  4. `280^@`
Show Answers Only

`C`

Show Worked Solution

 

`/_ ODT` `=90^@\ \ text{(line through centre bisecting chord)}`
`/_OAT` `= 90^@\ \ text{(tangent ⊥ to radius at point of contact)}`
`/_ DOA` `= 360-(90 + 90 + 40)`
  `= 140^@`

 

`:. DOA\ \ text{(reflex)}` `= 360-140`
  `= 220^@`

`=>   C`

Plane Geometry, EXT1 2015 HSC 3 MC

Two secants from the point `P` intersect a circle as shown in the diagram.
  

What is the value of `x`?

  1. `2`
  2. `5`
  3. `7`
  4. `8`
Show Answers Only

`B`

Show Worked Solution

`text{Property: products of intercepts of secants from external point are equal}`

`x(x + 3)` `= 4(4 + 6)`
`x^2 + 3x` `= 40`
`x^2 + 3x-40` `= 0`
`(x-5)(x + 8)` `= 0`

 
`:.x = 5,\ \  (x>0)`

`=>B`

Plane Geometry, EXT1 2014 HSC 1 MC

The points \(A\), \(B\) and \(C\) lie on a circle with centre \(O\), as shown in the diagram.

The size of \(\angle ACB\) is 40°.

 What is the size of \(\angle AOB\)?

  1. \(20^{\circ}\)
  2. \(40^{\circ}\)
  3. \(70^{\circ}\)
  4. \(80^{\circ}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\angle AOB = 2 \times 40 = 80^{\circ}\)

\(\text{(angles at centre and circumference on arc}\ AB\text{)}\) 

\(\Rightarrow D\)

Plane Geometry, EXT1 2012 HSC 10 MC

The points `A`, `B` and `P` lie on a circle centred at `O`. The tangents to the circle at `A` and `B` meet at the point `T`, and `/_ATB = theta`.

 What is `/_APB` in terms of  `theta`? 

  1. `theta/2`  
  2. `90^@-theta/2`
  3. `theta` 
  4. `180^@-theta` 
Show Answers Only

`B`

Show Worked Solution

`/_ BOA= 2 xx /_ APB`

`text{(angles at centre and circumference on arc}\ AB text{)}`

`/_TAO = /_ TBO = 90^@\ text{(angle between radius and tangent)}`

`:.\ theta + /_BOA` `= 180^@\ text{(angle sum of quadrilateral}\ TAOB text{)}`
`theta + 2 xx /_APB` `= 180^@`
`2 xx /_APB` `= 180^@-theta`
`/_APB` `= 90^@-theta/2`

 
`=>  B`