Circle Geometry (1 Viewer)

vds700 · Mar 1, 2009

Hi all, a friend asked me to help her with some circle geo questions, i was able able to do some, but not all. Appreciate some help with these.

1.
a) calculate the distance of the chord of length 24cm from the centre of a circle of radius 13cm- Done, answer = 5

b) the length of the tangents drawn from an external point to the extremities of the chord

17. two circles of radii 5cm and 8cm touch each other externally, calculate the length of the common tangent

18. If the radii of two intersecting circles are 17cm and 10cm and the length of the common chord if 16cm, calculate A) the length of the line joining their centres B) the length of the common tangent.

19. if the radii of two intersecting circles are 51cm and 74cm and the length of the common chord is 48cm, calculate A)the length of the line joining their centres B) the length of the common tangent

Thanks!

kurt.physics · Mar 1, 2009

vds700 said:
17. two circles of radii 5cm and 8cm touch each other externally, calculate the length of the common tangent

Thanks!

17.

(Here is the diagram)

The common tangent is the tangent that is common to both.

The basic idea behind this working is that we made a rectangle (this is possible because the radius makes a right angle with the tangent), then we use pythagoras to find the bottom side of the rectangle, thus finding the common tangent (as opposite sides of a rectangle are equal).

$the \,\,\, common \,\,\, tangent \,\,\, equals \,\,\, \sqrt{3^2 + (8+5)^2} = \fbox{\sqrt{178} }$

kurt.physics · Mar 1, 2009

vds700 said:
18. If the radii of two intersecting circles are 17cm and 10cm and the length of the common chord if 16cm, calculate A) the length of the line joining their centres B) the length of the common tangent.

18. The common chord is the red line.

Basically its another use of Pythagoras. We know that when two circles intersect, the line passing through their centres bisect the common chord (you can draw the fliped triangle above and use congruency to prove this). Also, the common chord makes right angles to the line connecting the centres.

$A) \sqrt{17^2 - 8^2} + \sqrt{10^2 - 8^2}$

$we \,\,\, are \,\,\, finding \,\,\, both \,\,\, sides \,\,\, of \,\,\, the \,\,\, line \,\,\, connecting \,\,\, the \,\,\, centres \,\,\, and \,\,\, adding \,\,\, them \,\,\, together.$

$So, \,\,\, \sqrt{17^2 - 8^2} + \sqrt{10^2 - 8^2} = 15 + 6$

$= \fbox{21cm}$

$B, \,\,\, We \,\,\, use \,\,\, the \,\,\, same \,\,\, method \,\,\, we \,\,\, used \,\,\, in \,\,\, the \,\,\, previous \,\,\, question \,\,\, about \,\,\, common \,\,\, tangents.$

$\sqrt{7^2 + 21^2} = \sqrt{490} = \fbox{7\sqrt{10}cm }$

kurt.physics · Mar 1, 2009

i'll leave the last one to you ;P

if you have trouble, im happy to help

vds700 · Mar 1, 2009

Thanks so much

cindyluu2 · Jan 20, 2012

Can someone please do question 1?

Sainteced · Jan 20, 2012

Kurt, those are some very nice diagrams that you have up there.

May I ask with what program are you making them?

IamBread · Jan 20, 2012

cindyluu2 said:
Can someone please do question 1?

This is going to be hard on a computer, but I'll see what I can do..

First, imagine a circle (or draw) of radius 13cm. This circle has a chord of 24cm, meaning that it connects to points on the circumference of the circle, and the distance between them is 24cm. Now you are asked to find the distance from the center to the chord, which is going to make a line perpendicular to the chord coming from the center, bisecting the chord. If we then join a line from the center to one of the points where the chord touches the circumference of the circle, we get a right angled triangle with sides 13, 12 and the short side unknown. Using Pythagoras we find it is equal to 5 cm.

Hope that makes sense.

cindyluu2 · Jan 20, 2012

Thank you so much, 'IamBread' Is it okay if you help me with part b too?

SpiralFlex · Jan 20, 2012

$First of all always draw a pretty diagram.$

$I can deduce the following from my diagram. Pretty diagram:$

${\color{Purple} \bullet} \angle EAO=90^\circ $\;(A tangent draw to a point of contact will make an angle of 90 with the radius)$$

${\color{Purple} \bullet} $ Similarly,$\; \angle EBO=90^\circ$

${\color{Purple} \bullet} $ I also know, $ AE=BE\; $(Two tangents from an external point have equal lengths.)$$

${\color{Purple} \bullet}$ Conclusively if the above case was true, then I know$ \;\Delta ABE\; $is actually an isosceles triangle.$$

$Let's apply our observations.$

$$From$\; \Delta AOD$

$\angle OAD = \sin^{-1} \frac{5}{13}$

$\therefore \angle OAD= 22.62^\circ$

$$We have said that$\; \angle EAO=90^\circ$

$So that means$ \; \angle EAD=90-22.62^\circ $\;(Complementary angles.)$

$\therefore \; \angle EAD = 67.38^\circ$

$As deduced before $\Delta ABE$ is actually an isosceles triangle.$

$This means I can find the lengths of AE and BE easily by using the sine rule.$

$\angle AEB=45.24^\circ $\;(Interior angle sum of a triangle.)$$

$By using the sine rule,$

$\frac{\sin 45.24^\circ }{24}=\frac{\sin 67.38^\circ }{AE}$

$\therefore AE=31.2 $\;cm$$

$So it follows, $ BE=31.2 $\;cm$

IamBread · Jan 20, 2012

cindyluu2 said:
Thank you so much, 'IamBread' Is it okay if you help me with part b too?

I'll let Spiralflex do that, he uses pretty diagrams

SpiralFlex · Jan 20, 2012

IamBread said:
I'll let Spiralflex do that, he uses pretty diagrams

Wait until 2013, I will blow everyone's minds with more awesome diagrams.

jenslekman · Jan 20, 2012

fitzpatrick 20c? and how do you draw like that D=

SpiralFlex · Jan 20, 2012

jenslekman said:
fitzpatrick 20c? and how do you draw like that D=

Pardon?

IamBread · Jan 20, 2012

SpiralFlex said:
Wait until 2013, I will blow everyone's minds with more awesome diagrams.

Hehe looking forward to it!

cindyluu2 · Jan 20, 2012

THANK YOU SO MUCH SPIRALFLEX!!

SpiralFlex · Jan 20, 2012

cindyluu2 said:
THANK YOU SO MUCH SPIRALFLEX!!

Anything else human?

Circle Geometry (1 Viewer)

vds700

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