Turn on thread page Beta

As core 1 Equation of a circle- Sixth form watch

    • Thread Starter
    Offline

    10
    ReputationRep:
    Okk i dont see where ive gone wrong, if someone could help?

    the question is
    1). P(6,7) is a point on a circle with its centre at (-3,1), find the equation of the normal to circle at P

    So i found the gradient first and i got 2/3, i then put it into the forumla

    y-7=2/3(x-6) i then x3 to get rid of the fraction and got 3(y-7)=2(x-6) and expanded it to get 3y-21=2x-12 and rearranged it to make y the subject and got the answer:

    y=2/3x +9 but the asnwer book said it was y=2/3x+3

    what did i do wrong?
    Offline

    0
    ReputationRep:
    3y-21 = 2x-12
    3y = 2x +9
    y= 2/3 x + 3
    Offline

    14
    ReputationRep:
    Right, well you know the general formula of a circle is:

    (x-a)^2+(y-b)^2=r^2

    So to begin with you have:

    (x+3)^2+(y-1)^2=r^2

    Then you need to find the distance between the two points, giving the radius of the circle. You just need phythagoras here:

    Root: (x2-x1)^2+(y2-y1)^2

    And as that is r^2, you don't need to root that at all.

    So your final equation is:

    (x+3)^2+(y-1)^2=127

    I think, correct me if am wrong
    Offline

    14
    ReputationRep:
    (Original post by AspiringGenius)
    Right, well you know the general formula of a circle is:

    (x-a)^2+(y-b)^2=r^2

    So to begin with you have:

    (x+3)^2+(y-1)^2=r^2

    Then you need to find the distance between the two points, giving the radius of the circle. You just need phythagoras here:

    Root: (x2-x1)^2+(y2-y1)^2

    And as that is r^2, you don't need to root that at all.

    So your final equation is:

    (x+3)^2+(y-1)^2=127

    I think, correct me if am wrong
    Almost, 81 + 36 = 117 not 127

    Also latex helps a lot in answering questions - I was a bit overwhelmed by it at first but it is really easy to use when you get used to it.


    EDIT: But more importantly I just realised you didn't answer the question at all It was asking the equation of the normal to the tangent at point P.
    Offline

    14
    ReputationRep:
    (Original post by hassi94)
    Almost, 81 + 36 = 117 not 127

    Also latex helps a lot in answering questions - I was a bit overwhelmed by it at first but it is really easy to use when you get used to it.
    Damn. I hate you basic arithmetic!
    Offline

    14
    ReputationRep:
    (Original post by Gary)
    Okk i dont see where ive gone wrong, if someone could help?

    the question is
    1). P(6,7) is a point on a circle with its centre at (-3,1), find the equation of the normal to circle at P

    So i found the gradient first and i got 2/3, i then put it into the forumla

    y-7=2/3(x-6) i then x3 to get rid of the fraction and got 3(y-7)=2(x-6) and expanded it to get 3y-21=2x-12 and rearranged it to make y the subject and got the answer:

    y=2/3x +9 but the asnwer book said it was y=2/3x+3

    what did i do wrong?
    Once you have the gradient of 2/3 you have:

     y = \frac{2}{3}x + c

x = 6, y = 7



7 = \frac{2}{3} (6) + c

7 = 4 + c

c = 3

    Hence,
    

y = \frac{2}{3}x + 3
    Offline

    0
    ReputationRep:
    is it just me or is the gradient of a normal -1/m

    and the gradient here is 2/3 => gradient of normal is -3/2 ??? or am i just being silly
    Offline

    14
    ReputationRep:
    (Original post by meraphox)
    is it just me or is the gradient of a normal -1/m

    and the gradient here is 2/3 => gradient of normal is -3/2 ??? or am i just being silly
    The gradient of 2/3 was worked out as the line from the centre to point P, which is the same as the gradient of the normal (and exactly the same line for that matter).
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 27, 2011

University open days

  • University of Chichester
    Multiple Departments Undergraduate
    Thu, 25 Oct '18
  • Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 26 Oct '18
  • University of Lincoln
    Open Day Undergraduate
    Sat, 27 Oct '18
Poll
Who do you think it's more helpful to talk about mental health with?
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

Equations

Best calculators for A level Maths

Tips on which model to get

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.