Extreme maths challenge (1 Viewer)

Real Madrid

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A circle has the centre C(-1,3) and radius 5 units

1. The line 3x-y+1=0 meets the circle at two points. Find their co-ordinates.

2. Let the coordinates be X and Y, where Y is the coordinate directly below the centre C. Find the coordinates of point Z where YZ is a diameter of the circle.

3. Hence show that AngleZXY = 90 degrees.
 

bubblesss

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1) first you find the equation of the circle which is x^2 +y^2 +2x-6y-15=0
then you solve the two equations( the circle and the line) to find the coordinates of x znd y. after solving u'll get x (2,7) and y(-1, -2)
 

bubblesss

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Real Madrid said:
A circle has the centre C(-1,3) and radius 5 units

1. The line 3x-y+1=0 meets the circle at two points. Find their co-ordinates.

2. Let the coordinates be X and Y, where Y is the coordinate directly below the centre C. Find the coordinates of point Z where YZ is a diameter of the circle.

3. Hence show that AngleZXY = 90 degrees.
for 2) use midpoint theorem

x1 + x2/2 ; y1+y2/2 = -1;3
x1 - 1/2 = -1 y1 - 2= 3
x1 = -1/2 y1 = 5

3) since yz is the diameter then angle zxy is 90 degrees ( angle in the semi-circle)
 

moshizzle

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1. (2, 7) and (-1, -2) as stated by bubblesss :p

2. so X = (2, 7) and Y = (-1, -2)
if YZ is a diameter, then its distance would be 10 units.
so plug into distance formula and get an equation.
right..? and then..? :uhoh: haha.
 

moshizzle

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bubblesss said:
for 2) use midpoint theorem

x1 + x2/2 ; y1+y2/2 = -1;3
x1 - 1/2 = -1 y1 - 2= 3
x1 = -1/2 y1 = 5

3) since yz is the diameter then angle zxy is 90 degrees ( angle in the semi-circle)
oh - i see :p
:rofl:
 

lyounamu

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Real Madrid said:
A circle has the centre C(-1,3) and radius 5 units

1. The line 3x-y+1=0 meets the circle at two points. Find their co-ordinates.

2. Let the coordinates be X and Y, where Y is the coordinate directly below the centre C. Find the coordinates of point Z where YZ is a diameter of the circle.

3. Hence show that AngleZXY = 90 degrees.
1. Done by bubbless. I think that's more concise and effective.

2. YZ must be 10 with Y 5 units below the centre and Z 5 units above the centre. So Z = (-1,8)

3. Z and Y are the points below and above the centre (i.e. forming a diameter). Therefore, the semi-circle is formed. Any angle in a semi-circle is 90 degrees.

By the way, stop naming the threads ridiculously. Those questions weren't that difficult at all as with other questions you posted up. People will still respond to you even if the questions are ridiculously easy.
 
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munchiecrunchie

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lyounamu said:
By the way, stop naming the threads ridiculously. Those questions weren't that difficult at all as with other questions you posted up. People will still respond to you even if the questions are ridiculously easy.
Its probably an attention thing, you know, more people are likely to respond to the post if its some sort of a challenge, and hence you get what you want (a solution) quicker.
 

Real Madrid

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People here like challenges?

Can someone show their working for 1?
 
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bubblesss

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Real Madrid said:
People here like challenges?

Can someone show their working for 1?
x^2 + y^2 +2x-6y-15=0.....................(1)
3x-y+1=0......................(2)

from 2 find x
3x-y+1=0
y= 1+3x.....................(3)

sub 3 in 1 and u'll get the values. u can do the subbing part rite?
 

bubblesss

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Real Madrid said:
When i sub it goes all wrong...

I have no idea why im wrong
subbing 3 in 1 we get

x^2+ (1+3x)^2 +2x-6(1+3x)-15=0
10x^2-10x-20=0
/ by 10, x^2 - x -2=0

(x-2)(x+1)=0
x=2 or -1
when x=2 , y=1+3x =1+6=7........P co ordinates
when x=-1, y= 1-3= -2 Qcoordinates
 

ratcher0071

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Real Madrid said:
A circle has the centre C(-1,3) and radius 5 units

1. The line 3x-y+1=0 meets the circle at two points. Find their co-ordinates.

2. Let the coordinates be X and Y, where Y is the coordinate directly below the centre C. Find the coordinates of point Z where YZ is a diameter of the circle.

3. Hence show that AngleZXY = 90 degrees.
1)
Equation of Circle:
(x+1)^2 + (y-3)^2 = 25 ...1

Equation of line:
3x-y+1=0 ...2

Rearrange 2 for y
y=3x+1 ...3

SUB 3 into 1
(x+1)^2 + ((3x+1)-3)^2 = 25
x^2 + 2x+1 + 9x^2 - 12x +4 =25
10x^2 -10x -20=0
x^2 -x-2=0
(x-2)(x+1)=0
x=2 or x=-1

SUB x values into 3
y=3(2)+1
y=7

y=3(-1)+1
y=-2

therefore points are Y(-1,-2) and X(2,7)

2) point Z(-1,8)

3) therefore AngleZXY=90 degrees (YZ is diameter, angle in a semicircle is 90 degrees)
 

tommykins

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I have to admit I'm solely disappointed at your threads being labelled "maths challenge", really am.
 

youngminii

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tommykins said:
I have to admit I'm solely disappointed at your threads being labelled "maths challenge", really am.
Don't say that
It makes me feel smart when I solve them, lol
 

SeftonIsAHole

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Real Madrid said:
Having trouble on 1 roflmao

I always get values of 2,-1 for x
you're on the right track... now just sub in those values back into 3x-y+1=0
to get your co-ordinates.
 

jamesboyd9

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bubblesss said:
for 2) use midpoint theorem

x1 + x2/2 ; y1+y2/2 = -1;3
x1 - 1/2 = -1 y1 - 2= 3
x1 = -1/2 y1 = 5
um, why did you do that?
this maths challenge isn't very extreme... I was disappointed. wait till you get to ext2 maths
 

Cookie182

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Lol to the OP

ask your teacher!

Its so obvious that you ahve no idea and hence post it just to copy the working/answer.
 

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