Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    3
    ReputationRep:
    https://a44694f152ee562e38a9454969d2...%20Edexcel.pdf

    question 5 a i have no idea how to do this help pls

    Zacken
    Offline

    22
    ReputationRep:
    (Original post by thefatone)
    https://a44694f152ee562e38a9454969d2...%20Edexcel.pdf

    question 5 a i have no idea how to do this help pls

    Zacken
    GCSE theorem, angle in a semi-cricle is a right-angle.

    Here PR is a diameter so PQ and QR form an angle in a semi-circle, making an angle of 90 degrees, hence PQ and QR are perpendicular. What can you then say about their gradients? More specficially, their product of gradients. That'll give you an equation in a which you can then solve.
    Offline

    2
    ReputationRep:
    Calculate the gradient of PQ and then as QR is perpendicular the the gradient of QR will be -3/2

    Then work out the equation of the line QR by susbstituting in the coordinate of Q to get the equation

    Plug in y = 4 and calculate a to be 13
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    GCSE theorem, angle in a semi-cricle is a right-angle.

    Here PR is a diameter so PQ and QR form an angle in a semi-circle, making an angle of 90 degrees, hence PQ and QR are perpendicular. What can you then say about their gradients? More specficially, their product of gradients. That'll give you an equation in a which you can then solve.
    i don't understand

    ......


    oh no i see ok right, i didn't see they were perpendicular right now i see thanks
    Offline

    22
    ReputationRep:
    (Original post by thefatone)
    i don't understand

    ......


    oh no i see ok right, i didn't see they were perpendicular right now i see thanks
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    we're getting there slowly all these silly things which i don't see -.-
    Offline

    18
    ReputationRep:
    a ) Diameter^2 : (a + 3)^2 + (4 - 2)^2 = a^2 + 6a + 9 + 4 = a^2 + 6a + 13

    (a - 9)^2 + ( 4 - 10)^2 = a^2 - 18a + 81 + 36 => a^2 - 18a + 117

    a^2 + 6a + 13 = 208 + a^2 - 18a + 117

    => 24a = 325 - 13 = 312

    a = 13

    b)Centre x = a + 3 / 2 = 8

    Centre y = 4 - 2 / 2 = 1

    Diameter^2 = 13^2 + 6 * 13 + 13 = 169 + 78 + 13 = 91 + 169 = 260

    Diameter = 16.12

    Radius = 8.06

    Radius^2 = 65

    (x - 8)^2 + (y - 1)^2 = 65
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Have you ever participated in a Secret Santa?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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