Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    Hi I'm studying from home so sometimes there are parts that my course omits from teaching me, at which point I defer to the internet, so hi!

    My question is referring to circles. Particularly the questions of the format:

    1) Write down the radius and the co-ordinates to the centre of:

    a) x2 + y2 + 3x - 5y + 2 = 0


    These I am fine with however I am stuck when the x2 and y2 terms are given coefficients,
    This is the exact question I am stuck on:

    b) 3x2 + 3y2 + 6x - 4y - 15 = 0

    Can someone please help?

    Thank you
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by Bysteven)
    Hi I'm studying from home so sometimes there are parts that my course omits from teaching me, at which point I defer to the internet, so hi!

    My question is referring to circles. Particularly the questions of the format:

    1) Write down the radius and the co-ordinates to the centre of:

    a) x2 + y2 + 3x - 5y + 2 = 0


    These I am fine with however I am stuck when the x2 and y2 terms are given coefficients,
    This is the exact question I am stuck on:

    b) 3x2 + 3y2 + 6x - 4y - 15 = 0

    Can someone please help?

    Thank you
    Try dividing the equation by 3
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Bysteven)
    ...
    The coefficients of x^2 and y^2 will always be the same on circles questions so notnek's method is guaranteed to work every time.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Mr M)
    The coefficients of x^2 and y^2 will always be the same on circles questions so notnek's method is guaranteed to work every time.
    Thanks guys, I divided by 3 however my answer for the radius differs from the given answer at the back of the book.
    My answer was 7 25/36 the textbook is saying 1/3√58.


    Here is how I got there:

    - 3x2 + 3y2 + 6x - 4y - 15 = 0

    - x2 + 3x + y2 - 4/3 y - 5 = 0

    - (x + 1.5)2 + (y - 4/6)2 - 2.25 - 16/36 - 5 = 0

    - (x + 1.5)2 + (y - 2/3)2 = 7.25 + 8/18 ( or 7 + 1/4 + 8/18)

    - (x + 1.5)2 + (y - 2/3)2 = √ (7 50/72) or √ (7 25/36)

    The Co-ordinates are right at (-1,2/3) but can someone please explain how I got the radius wrong.

    Thank you!
    • Study Helper
    Offline

    9
    ReputationRep:
    Study Helper
    (Original post by Bysteven)
    Thanks guys, I divided by 3 however my answer for the radius differs from the given answer at the back of the book.
    My answer was 7 25/36 the textbook is saying 1/3√58.


    Here is how I got there:

    - 3x2 + 3y2 + 6x - 4y - 15 = 0

    - x2 + 3x + y2 - 4/3 y - 5 = 0

    - (x + 1.5)2 + (y - 4/6)2 - 2.25 - 16/36 - 5 = 0

    - (x + 1.5)2 + (y - 2/3)2 = 7.25 + 8/18 ( or 7 + 1/4 + 8/18)

    - (x + 1.5)2 + (y - 2/3)2 = √ (7 50/72) or √ (7 25/36)

    The Co-ordinates are right at (-1,2/3) but can someone please explain how I got the radius wrong.

    Thank you!
    6x divided by 3 is 2x NOT 3x!!
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by brianeverit)
    6x divided by 3 is 2x NOT 3x!!
    Thanks I can't believe I missed that,
    In which case,


    - 3x2 + 3y2 + 6x - 4y - 15 = 0

    - x2 + 2x + y2 - 4/3 y - 5 = 0

    - (x + 1)2 + (y - 4/6)2 - 1 - 16/36 - 5 = 0

    - (x + 1)2 + (y - 2/3)2 = 6 16/36

    - (x + 1)2 + (y - 2/3)2 = √ 6 4/9

    Is this what everyone else would have solved it at? In which case I am still off the final answer.
    Thanks so much for all the help
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Bysteven)
    Thanks I can't believe I missed that,
    In which case,


    - 3x2 + 3y2 + 6x - 4y - 15 = 0

    - x2 + 2x + y2 - 4/3 y - 5 = 0

    - (x + 1)2 + (y - 4/6)2 - 1 - 16/36 - 5 = 0

    - (x + 1)2 + (y - 2/3)2 = 6 16/36

    - (x + 1)2 + (y - 2/3)2 = √ 6 4/9

    Is this what everyone else would have solved it at? In which case I am still off the final answer.
    Thanks so much for all the help
    Actually I've just worked it out that 1/3 √58 AND √6 4/9 are both equal to 2.53859 which means I have got the correct answer

    Would I still get the mark or would it have to match the format of the given answer?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Bysteven)
    Actually I've just worked it out that 1/3 √58 AND √6 4/9 are both equal to 2.53859 which means I have got the correct answer

    Would I still get the mark or would it have to match the format of the given answer?
    Exact form is fine.
    • Study Helper
    Offline

    9
    ReputationRep:
    Study Helper
    (Original post by Bysteven)
    Actually I've just worked it out that 1/3 √58 AND √6 4/9 are both equal to 2.53859 which means I have got the correct answer

    Would I still get the mark or would it have to match the format of the given answer?
    You must be careful with your notation , writing  \sqrt6\frac{4}{9} it is not clear that the fraction is included in the square root. Writing 6\frac{4}{9} as an improper fraction, i.e. \frac{58}{9} is much better and we then have \sqrt{\frac{58}{9}}=\frac{\sqrt{  58}}{3} as required
    • Thread Starter
    Offline

    1
    ReputationRep:
    Gotcha thanks alot this helped loads )
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: August 17, 2014
Poll
Do I go to The Streets tomorrow night?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.