Turn on thread page Beta

p3 help watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hi,

    Could you please help me with these circle questions?

    1. Show that the line with eqn. 2x-3y+26=0 is a tangent to the circle with the eqn. x^2 + y^2 - 4x + 6y - 104 = 0. The point of contact is T, find the co-ordinates of the point on the circle which is diametrically opposite to T.

    2. Line with eqn. y=mx+c is tangent to the circle with eqn. x^2+y^2-6x-6y+17=0.
    Find the possible values of m.

    3. Prove that the circle with eqn. x^2+y^2-2ax-2by+b^2=0 touches the y-axis. Hence, or otherwise, find the eqns. of the 2 circles which pass through the points (1,2) and (2,3) and which touch the y-axis.
    Offline

    0
    ReputationRep:
    find y in terms of x in the linear eq, then sub that in to the circle, to find Y, then sub y in to the linear to find T.
    since the linears gradient is 2/3, and is a tangent, so, the radius that it meets has a gradient of -3/2
    use y-y1=m(x-x1) to find eq of the radius, then, set that equation equal to the circle, which should find T, and the other point, diametrically opposite to T

    this is how i would attempt it, please bear in mind, im a C2 student, who, for an hour a week, reads P3
    • Thread Starter
    Offline

    0
    ReputationRep:
    anyone?
    Offline

    0
    ReputationRep:
    i tried subbing in x=1.5y-13 but it doesnt seem to work
    Offline

    1
    ReputationRep:
    (Original post by matt.forster)
    find y in terms of x in the linear eq, then sub that in to the circle, to find Y, then sub y in to the linear to find T.
    since the linears gradient is 2/3, and is a tangent, so, the radius that it meets has a gradient of -3/2
    use y-y1=m(x-x1) to find eq of the radius, then, set that equation equal to the circle, which should find T, and the other point, diametrically opposite to T

    this is how i would attempt it, please bear in mind, im a C2 student, who, for an hour a week, reads P3
    not bad just a few tweaks...

    find y in terms of x in the linear equation. rewrite the circle eq into proper circle form, ie (x-a)^2 + (y-b)^2 = (r^2). then sub the expression for y in terms of x and solve to show only one solution. put this value of x back into the linear equation to find the value of y.

    2x-3y+26=0
    y = (2x + 26)/3

    x^2 + y^2 - 4x + 6y - 104 = 0
    (x-2)^2 + (y+3)^2 = 117
    (x-2)^2 + [(2x + 26 + 9)/3]^2 = 117
    (x-2)^2 + (1/9)[2x + 35]^2 = 117
    (x^2) - 4x + 4 + (1/9)[4(x^2) + 140x + 1225] = 117
    (1/9)[(9x^2) - 36x + 36 + 4(x^2) + 140x + 1225] = 117
    (9x^2) - 36x + 36 + 4(x^2) + 140x + 1225 = 1053
    13(x^2) + 104x + 1261 = 1053
    13(x^2) + 104x + 208 = 0
    (x^2) + 8x + 16 = 0

    determinant = b^2 - 4ac
    = (8^2) - 4(1)(16)
    = 64 - 64
    = 0

    determinant = 0 => one repeated real root => tangent as only one common point...

    (x^2) + 8x + 16 = 0
    (x+4)(x+4) = 0
    x = -4

    y = (2x + 26)/3
    y = -8 + 26)/3
    y = 6
    Offline

    0
    ReputationRep:
    for the first one ive got it now, you have to sub in y=2/3x+26/3

    x^2 + (2/3x+26/3)^2 - 4x + 6(2/3x+26/3) - 104

    13/9x^2 + 104/9x + 208/9 = 0

    sqr root of b^2-4ac = 104^2 - 4 x 13 x 208 = 0

    this means its a tangent
    • Thread Starter
    Offline

    0
    ReputationRep:
    can you please answer 3, and i have another question, if your given the eqn of a circle, (x-a)^2 + (y-b)^2 = r^2. and the line OT is a tangent to the circle.
    how do you find an eqn. of the line OT.
    Offline

    15
    ReputationRep:
    (Original post by manps)
    can you please answer 3, and i have another question, if your given the eqn of a circle, (x-a)^2 + (y-b)^2 = r^2. and the line OT is a tangent to the circle.
    how do you find an eqn. of the line OT.
    Use pythagoras as the tangent makes a right angle with a radius.
    Offline

    10
    ReputationRep:
    3.
    If the circle touches the y-axis, then there's only 1 point of intersection between it and the line x=0. That is:
    x² + y² - 2ax - 2by + b² = 0
    y² - 2by + b² = 0

    D = B² - 4AC = (2b)² - 4(1)(b²) = 0, hence there is only one point of contact. QED.

    Now let's plug (1,2) and (2,3) into the equation of the circle. Upon simplifcation we get:
    b² - 4b - 2a + 5 = 0
    b² - 6b - 4a + 13 = 0

    Eliminating a:
    b² - 2b - 3 = 0
    (b-3)(b+1) = 0
    b=3 or b=-1

    Now find the corresponding values of a, and then you can use the pair of values of a & b to find the two equations.
 
 
 
Turn on thread page Beta
Updated: May 21, 2005

University open days

  • University of East Anglia
    All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
    Wed, 30 Jan '19
  • Aston University
    Postgraduate Open Day Postgraduate
    Wed, 30 Jan '19
  • Solent University
    Careers in maritime Undergraduate
    Sat, 2 Feb '19
Poll
Brexit: Given the chance now, would you vote leave or remain?
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.