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

    9
    ReputationRep:
    the line with equation y=3x-2 does not intersect the circle with centre (0,0) and radius r . find the range of possible values of r. Answer is 0<r< root 2/5
    So far I've done substituted 0 for x and then for y. Whereby: (0,-2) and (2/3,0)
    I've used the formula (x-a)^2 +(y-b)^2=r^2
    (0-2/3)^2+ (-2,0) ^2 =r^2
    4/9+4=r^2
    40/9=r^2
    2 root 10 divided by 3
    But this is definitely not correct
    Where have I gone wrong

    Thanks
    Offline

    15
    ReputationRep:
    Set up an equation as if you were trying solve the simultaneous equations. Then you have a quadratic which I will have no solutions if the curves don't cross - think discriminant.
    Offline

    11
    ReputationRep:
    Find the shortest distance between the line and origin. That will be the maximum radius of the circle
    Offline

    21
    ReputationRep:
    (Original post by Musicanor)
    the line with equation y=3x-2 does not intersect the circle with centre (0,0) and radius r . find the range of possible values of r. Answer is 0<r< root 2/5
    I'd tackle this slightly differently (I do that a lot).

    The point on the line closest to the origin is its point of intersection with a line perpendicular to it, and through the origin.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by Laissez‒faire)
    Find the shortest distance between the line and origin. That will be the maximum radius of the circle
    2/3???
    Offline

    21
    ReputationRep:
    (Original post by RogerOxon)
    The point on the line closest to the origin is its point of intersection with a line perpendicular to it, and through the origin.
    y=3x-2

    What is the equation of a line perpendicular to that, through the origin? (Hint: the product of the gradients of two perpendicular lines is -1)
    • Community Assistant
    • Welcome Squad
    Online

    20
    ReputationRep:
    (Original post by Musicanor)
    the line with equation y=3x-2 does not intersect the circle with centre (0,0) and radius r . find the range of possible values of r. Answer is 0<r< root 2/5
    So far I've done substituted 0 for x and then for y. Whereby: (0,-2) and (2/3,0)
    I've used the formula (x-a)^2 +(y-b)^2=r^2
    (0-2/3)^2+ (-2,0) ^2 =r^2
    4/9+4=r^2
    40/9=r^2
    2 root 10 divided by 3
    But this is definitely not correct
    Where have I gone wrong

    Thanks
    Take the line perpendicular to y=3x-2 through the origin. What is the point of intersection between these two lines? Hence what is the distance between the origin and this point?

    Get it?
    Offline

    11
    ReputationRep:
    (Original post by Musicanor)
    2/3???
    I haven't done it. You need to find an eq of a line perpendicular which goes through 0,0 and then find their intersection. Then find the distance between that and the origin and you have the maximum radius.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by RogerOxon)
    I'd tackle this slightly differently (I do that a lot).

    The point on the line closest to the origin is its point of intersection with a line perpendicular to it, and through the origin.
    Thanks
    I've got:
    2/3--> -3/2
    Y=3(0) -3/2
    (0, -3/2)
    0=3x-3/2
    3/2= 3x
    3/2 / 3=X
    1/2=x

    (0-1/2)^2 + (-3/2 -0)^2=r^2
    1/4 +9/4 =r^2
    5/2=r^2
    Root 5/2 =r

    But answer is meant to be root 2/5
    ???
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by RogerOxon)
    y=3x-2

    What is the equation of a line perpendicular to that, through the origin? (Hint: the product of the gradients of two perpendicular lines is -1)
    -1/3
    Offline

    21
    ReputationRep:
    (Original post by Musicanor)
    -1/3
    Correct. So, the line perpendicular to y=3x-2 (and through the origin) is y=-\frac{x}{3}

    Now find the point of intersection of those lines.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by RogerOxon)
    Correct. So, the line perpendicular to y=3x+2 (and through the origin) is y=-\frac{x}{3}

    Now find the point of intersection of those lines.
    Y=3x+2
    Y=-x/3
    Y-y1= m(x-X1)
    Y-2= -x/3 (x-0)
    Y-2= -x/3 x
    Y= -x/3 +2

    3(3x+2)=-x
    9x +6=-x
    10x+6=0
    2(5x+3)=0
    5x+3=0
    X=-3/5
    Is this close:
    2/3--> -3/2 Y=3(0) -3/2 (0, -3/2) 0=3x-3/2 3/2= 3x 3/2 / 3=X 1/2=x (0-1/2)^2 + (-3/2 -0)^2=r^2 1/4 +9/4 =r^2 5/2=r^2 Root 5/2 =r
    Thank you so much for helping me
    Offline

    21
    ReputationRep:
    (Original post by Musicanor)
    Y-y1= m(x-X1)
    Y-2= -x/3 (x-0)
    Y-2= -x/3 x
    Y= -x/3 +2
    I don't follow this.

    3(3x+2)=-x
    9x +6=-x
    10x+6=0
    2(5x+3)=0
    5x+3=0
    X=-3/5
    Sorry - I misquoted the first line - it's 3x-2 (I'd put +), which gives x=+\frac{3}{5} and y=\frac{-1}{5}.

    By Pythagoras, we get d=\sqrt{(\frac{3}{5})^2+(\frac{1  }{5})^2}=\sqrt{\frac{10}{25}}= \sqrt{\frac{2}{5}}
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by RogerOxon)
    I don't follow this.


    Sorry - I misquoted the first line - it's 3x-2 (I'd put +), which gives x=+\frac{3}{5} and y=\frac{-1}{5}.

    By Pythagoras, we get d=\sqrt{(\frac{3}{5})^2+(\frac{1  }{5})^2}=\sqrt{\frac{10}{25}}=\s  qrt{\frac{2}{5}}
    Yay Thanks I understand now
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by RogerOxon)
    I don't follow this.


    Sorry - I misquoted the first line - it's 3x-2 (I'd put +), which gives x=+\frac{3}{5} and y=\frac{-1}{5}.

    By Pythagoras, we get d=\sqrt{(\frac{3}{5})^2+(\frac{1  }{5})^2}=\sqrt{\frac{10}{25}}= \sqrt{\frac{2}{5}}
    How did you get 3/5 and 1/5?
    Offline

    21
    ReputationRep:
    (Original post by Musicanor)
    How did you get 3/5 and 1/5?
    The first line is y=3x-2

    The product of the gradients of perpendicular lines is -1. The line perpendicular to this, and through the origin, is therefore y=\frac{-x}{3}

    These lines intersect when:

    \frac{-x}{3}=3x-2

    \therefore \frac{10x}{3}=2

    \therefore x=\frac{3}{5}

    We can get the y coordinate by using that x value in the equation of either line:

    \therefore y=\frac{-x}{3}=\frac{-\frac{3}{5}}{3}=\frac{-3}{15}=\frac{-1}{5}

    The distance from the origin to the point of intersection i the sqare roto of the sum of the squares of the difference in x and the difference in y - from the origin, it's just the squares of the coordinates. When I squared the \frac{-1}{5}, I dropped the minus sign, as it doesn't make any difference.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by RogerOxon)
    The first line is y=3x-2

    The product of the gradients of perpendicular lines is -1. The line perpendicular to this, and through the origin, is therefore y=\frac{-x}{3}

    These lines intersect when:

    \frac{-x}{3}=3x-2

    \therefore \frac{10x}{3}=2

    \therefore x=\frac{3}{5}

    We can get the y coordinate by using that x value in the equation of either line:

    \therefore y=\frac{-x}{3}=\frac{-\frac{3}{5}}{3}=\frac{-3}{15}=\frac{-1}{5}

    The distance from the origin to the point of intersection i the sqare roto of the sum of the squares of the difference in x and the difference in y - from the origin, it's just the squares of the coordinates. When I squared the \frac{-1}{5}, I dropped the minus sign, as it doesn't make any difference.
    Thank you so so so sooooo much. You've helped with a lot you're the best
 
 
 
  • 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
    What newspaper do you read/prefer?
    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.