Prelim 2016 Maths Help Thread (2 Viewers)

[davidgoes4wce]

Please help me thanks...

Domain and range answered by Integrand

 

[sgtgummybear]

A circle has centre C(-1, 3) and
radius 5 units.
(a) Find the equation of the circle
(b) The line 3x — y + 1 = O meets
the circle at two points. Find
their coordinates.
(c) 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.
(d) Hence show angleZXY = 90 degrees.

Please help

 

[eyeseeyou]

BTW Simorgh, you need to change the name of this thread to "HSC 2017 maths help thread"

 

[pikachu975]

A circle has centre C(-1, 3) and
radius 5 units.
(a) Find the equation of the circle
(b) The line 3x — y + 1 = O meets
the circle at two points. Find
their coordinates.
(c) 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.
(d) Hence show angleZXY = 90 degrees.

Please help

a)
Equation: (x-h)^2 + (y-k)^2 = r^2
so it is (x+1)^2 + (y-3)^2 = 25

b)
Use simultaneous equations by substitution
3x - y + 1 = 0
y = 3x + 1, sub this into the circle
(x+1)^2 + (3x-2)^2 = 25
x^2 + 2x + 1 + 9x^2 - 12x + 4 - 25 = 0
10x^2 - 10x - 20 = 0
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2, -1
sub these into the line equation to get y = 7, -2
so the answer is (2, 7) and (-1, -2)

c) Don't understand the question

Edit: typo

 

[sgtgummybear]

a)
Equation: (x-h)^2 + (y-k)^2 = r^2
so it is (x+1)^2 + (y-3)^2 = 25

b)
Use simultaneous equations by substitution
3x - y + 1 = 0
y = 3x + 1, sub this into the circle
(x+1)^2 + (3x-2)^2 = 25
x^2 + 2x + 1 + 9x^2 - 12x + 4 - 25 = 0
10x^2 - 10x - 20 = 0
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2, -1
sub these into the line equation to get y = 7, -2
so the answer is (2, 7) and (-1, -2)

c) Don't understand the question

Edit: typo

Hahah, thanks for clearing up the typo. I was getting so confused

Thank you so much for helping ^_^

 

[pikachu975]

Hahah, thanks for clearing up the typo. I was getting so confused

Thank you so much for helping ^_^

Do you have a diagram for part c so I can solve it? Thanks

Wait I just realised what the question meant lol

 

[pikachu975]

A circle has centre C(-1, 3) and
radius 5 units.
(a) Find the equation of the circle
(b) The line 3x — y + 1 = O meets
the circle at two points. Find
their coordinates.
(c) 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.
(d) Hence show angleZXY = 90 degrees.

Please help

c)
X(2,7) and Y(-1,-2) and Z(x,y)
So we know YZ passes through the diameter, ie YC = CZ so C is the midpoint of YZ
Using midpoint formula,
(-1 = (x-1)/2, 3 = (y-2)/2)
= (x = -1, y = 8)
Therefore Z(-1, 8)

d) Find gradients of line ZX and XY
m of ZX = (8-7)/(-1-2)
= -1/3
m of XY = (-2-7)(-1-2)
=
3

m1 x m2 = -1/3 x 3
= -1 hence they're perpendicular so angle ZXY = 90 degrees

 

[boredofstudiesuser1]

(c) 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.
(d) Hence show angleZXY = 90 degrees.

Please help

I am basing my answers off of Pikachu's for Part a and b

c) Y must be (-1,-2) (directly below c(-1-3))
If YZ are a diameter, Z is directly 10 units above Y, since the centre is between them
Therefore, Z(-1,8)

d) For ZXY to be 90 degrees, ZX must be perp. to XY
Eqn of XY is -y=-3x-1 (m=3)
Eqn of XZ is -3y=x-23 (m=-1/3)
If both gradients multiply to equal -1, then the lines are perpendicular
-1/3*3=-1
Therefore, they are perp. and ZXY is right angle (90 degrees)
There is a shortcut to this question:
Since ZY is a diameter and X is a point that lies on the circumference of the circle, ZXY has to be a right angle
But this is a circle theorem and you learn it in preliminary extension 1 (I assume these questions are just Preliminary Mathematics and they would rather you use the first method)

Hope this helps and that I got them right lol

 

[boredofstudiesuser1]

c)
X(2,7) and Y(-1,-2) and Z(x,y)
So we know YZ passes through the diameter, ie YC = CZ so C is the midpoint of YZ
Using midpoint formula,
(-1 = (x-1)/2, 3 = (y-2)/2)
= (x = -1, y = 8)
Therefore Z(-1, 8)

d) Find gradients of line ZX and XY
m of ZX = (8-7)/(-1-2)
= -1/3
m of XY = (-2-7)(-1-2)
=
3

m1 x m2 = -1/3 x 3
= -1 hence they're perpendicular so angle ZXY = 90 degrees

Lol yeah, you can find the gradients like that. I always forget and find the equation and then the gradient...

 

[sgtgummybear]

c)
X(2,7) and Y(-1,-2) and Z(x,y)
So we know YZ passes through the diameter, ie YC = CZ so C is the midpoint of YZ
Using midpoint formula,
(-1 = (x-1)/2, 3 = (y-2)/2)
= (x = -1, y = 8)
Therefore Z(-1, 8)

d) Find gradients of line ZX and XY
m of ZX = (8-7)/(-1-2)
= -1/3
m of XY = (-2-7)(-1-2)
=
3

m1 x m2 = -1/3 x 3
= -1 hence they're perpendicular so angle ZXY = 90 degrees

Awesome. Once again, thank you.

 

[sgtgummybear]

I am basing my answers off of Pikachu's for Part a and b

c) Y must be (-1,-2) (directly below c(-1-3))
If YZ are a diameter, Z is directly 10 units above Y, since the centre is between them
Therefore, Z(-1,8)

d) For ZXY to be 90 degrees, ZX must be perp. to XY
Eqn of XY is -y=-3x-1 (m=3)
Eqn of XZ is -3y=x-23 (m=-1/3)
If both gradients multiply to equal -1, then the lines are perpendicular
-1/3*3=-1
Therefore, they are perp. and ZXY is right angle (90 degrees)
There is a shortcut to this question:
Since ZY is a diameter and X is a point that lies on the circumference of the circle, ZXY has to be a right angle
But this is a circle theorem and you learn it in preliminary extension 1 (I assume these questions are just Preliminary Mathematics and they would rather you use the first method)

Hope this helps and that I got them right lol

Oh, cool. More explanations! Thank you!

 

[mahdi24680]

I have the 2 u and 3 u Cambridge textbooks, and i dont know which one i should do. Should i go straight to the 3u book or do 2u and then 3u.

 

[KingOfActing]

I have the 2 u and 3 u Cambridge textbooks, and i dont know which one i should do. Should i go straight to the 3u book or do 2u and then 3u.

3U book encompasses 2u, you don't need the 2u book.

 

[eyeseeyou]

BTW to all 2017ers, what topic(s) do you start when you come back to school?

 

[jathu123]

BTW to all 2017ers, what topic(s) do you start when you come back to school?

2u: Geometric applications of calculus
3u: harder parametrics and circle geo/ext 1 curve sketching
4u: complex numbers

 

[eyeseeyou]

2u: Geometric applications of calculus
3u: harder parametrics and circle geo/ext 1 curve sketching
4u: complex numbers

Didn't you already do curve sketching in calculus?

 

[Orwell]

I start filling out my application to drop mathematics.

 

[pikachu975]

BTW to all 2017ers, what topic(s) do you start when you come back to school?

3u: Probably the miscellaneous stuff like inequalities, angles between lines, or circle geo
4u: Complex numbers probably

 

[eyeseeyou]

3u: Probably the miscellaneous stuff like inequalities, angles between lines, or circle geo
4u: Complex numbers probably

Is this after your HSC exams?

 

[pikachu975]

Is this after your HSC exams?

Yep, 3u and 4u don't start till week 3 because the HSC for maths is week 2 Friday.