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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)   

--- 6 WORK AREA LINES (style=lined) ---

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\(OB=5\)

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

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

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

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

Filed Under: Circle Geometry Tagged With: num-title-ct-path, smc-4240-50-Chord properties

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

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

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

Filed Under: Circle Geometry Tagged With: num-title-ct-patha, smc-4240-50-Chord properties

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

Filed Under: Circle Geometry Tagged With: num-title-ct-patha, smc-4240-50-Chord properties

Circle Geometry, SMB-005

In the diagram, two chords of a circle intersect. 

Find \(x\).  (2 marks) 
  

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\(x=8\)

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\(3 \times x\) \(=6 \times 4\ \ \text{(intercepts of intersecting chords)}\)  
\(x\) \(= \dfrac{24}{3} \)  
  \(=8\)  

Filed Under: Circle Geometry Tagged With: num-title-ct-patha, smc-4240-50-Chord properties

Circle Geometry, SMB-004

In the diagram, two chords of a circle intersect. 

Find \(x\).  (2 marks) 
  

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\(x=6\)

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\(7 \times x\) \(=3 \times 14\ \ \text{(intercepts of intersecting chords)}\)  
\(x\) \(= \dfrac{42}{7} \)  
  \(=6\)  

Filed Under: Circle Geometry Tagged With: num-title-ct-patha, smc-4240-50-Chord properties

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)

--- 4 WORK AREA LINES (style=lined) ---

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\(\text{See Worked Solutions}\)

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


 

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

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

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

Filed Under: Circle Geometry Tagged With: num-title-ct-patha, smc-4240-50-Chord properties

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^@`
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`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`

Filed Under: 2. Plane Geometry EXT1, Circle Geometry Tagged With: Band 4, num-title-ct-path, num-title-qs-hsc, smc-4240-50-Chord properties, smc-4240-60-Tangents

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