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

    11
    ReputationRep:
    There are two circles, Z and M. Circle Z has the equation (x - 1)^2 + (y - 1)^2 = 1, and circle M has the equation (x - 5)^2 + (y - 7)^2 = 9. Find the points that are the nearest to each other on both circles.

    I've worked out the distance from the center of Z to M, which is √54, and the gradient from the center of Z to the center of M is 3/2.

    I know I somehow need to apply the gradient to Z, and in reverse to M, to find the points that lie on the line ZM (center to center), which also lie on the edges of the circles. However, I haven't managed to find a way to do this.

    Could you please assist?
    Offline

    9
    ReputationRep:
    (Original post by UCASLord)
    There are two circles, Z and M. Circle Z has the equation (x - 1)^2 + (y - 1)^2 = 1, and circle M has the equation (x - 5)^2 + (y - 7)^2 = 9. Find the points that are the nearest to each other on both circles.

    I've worked out the distance from the center of Z to M, which is √54, and the gradient from the center of Z to the center of M is 3/2.

    I know I somehow need to apply the gradient to Z, and in reverse to M, to find the points that lie on the line ZM (center to center), which also lie on the edges of the circles. However, I haven't managed to find a way to do this.

    Could you please assist?
    One option is to find the equation of the line through the centres and then solve simultaneous equations to find the points where this line intersects the circles.

    A quicker option is to use vectors.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by BuryMathsTutor)
    One option is to find the equation of the line through the centres and then solve simultaneous equations to find the points where this line intersects the circles.

    A quicker option is to use vectors.
    I quickly worked this out for the (x - 1)^2 + (y - 1)^2 = 1 circle, and got x = 1 + 213/13, so from there y = 3/2(1 + 213/13) - 1/2... Would those be the right coordinates for the the point on the circle Z?
    Offline

    9
    ReputationRep:
    (Original post by UCASLord)
    I quickly worked this out for the (x - 1)^2 + (y - 1)^2 = 1 circle, and got x = 1 + 213/13, so from there y = 3/2(1 + 213/13) - 1/2... Would those be the right coordinates for the the point on the circle Z?
    That's correct. Don't forget to simplify the y value.
 
 
 
  • 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
    Did TEF Bronze Award affect your UCAS choices?
    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.