Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    The circle with equation x2 + y2 – 12x – 10y + k = 0 meets the coordinate axes at exactly three points.
    What is the value of k???
    • TSR Support Team
    • Clearing and Applications Advisor
    Offline

    21
    ReputationRep:
    TSR Support Team
    Clearing and Applications Advisor
    (Original post by Cameron45)
    The circle with equation x2 + y2 – 12x – 10y + k = 0 meets the coordinate axes at exactly three points.
    What is the value of k???
    From that you can work out the centre of the circle.
    Then, if it meets the coordinate axis exactly 3 times, this implies that it must meet once, and intersect twice. From the centre you should be able to tell the radius where it intersects twice and meets it once, you can use this to work out the value of k
    Let me know if you get stuck!
    • Thread Starter
    Offline

    2
    ReputationRep:
    Would the centre be (6,5)? Where do you go from there?
    (Original post by Lemur14)
    From that you can work out the centre of the circle.
    Then, if it meets the coordinate axis exactly 3 times, this implies that it must meet once, and intersect twice. From the centre you should be able to tell the radius where it intersects twice and meets it once, you can use this to work out the value of k
    Let me know if you get stuck!
    Offline

    20
    ReputationRep:
    because it meets the axes thrice, it follows that the radius must either equal the x coordinate of the centre or the y coordinate of the centre.
    Offline

    10
    ReputationRep:
    You can do it a different way - I think it's more about quadratic equation solving than circles.

    Where it meets the axis, either x or y must be zero.

    So, put x = 0, this will give you a quadratic in y (1)

    Similarly, putting y = 0 gives you a quadriatic in x. (2)

    Now, if there are only THREE places where it crosses the axes, then there are THREE solutions to those two quadratics which means one of them only has one solution. That means for one of them, the determinant is 0.

    So, I took my two new quadratic equations, found the value of k which gave a determinant of zero.

    That meant I had two possible values of k, one from (1) and one from (2)

    I put the value of k from (1) back into (2) to try to solve for x
    I put the value of (2) back into (1) to try to solve for y

    You don't actually have to find them, just check the determinants again.

    One of those two (not telling you which, try it) gave me a negative determinant which meant the equation would have NO solutions which means the original equation could not have THREE solutions.
    The other one woried. I found the markscheme, checked and it's right.

    Try it - if you can't get it to work out come back and ask again.
    • TSR Support Team
    • Clearing and Applications Advisor
    Offline

    21
    ReputationRep:
    TSR Support Team
    Clearing and Applications Advisor
    (Original post by Cameron45)
    Would the centre be (6,5)? Where do you go from there?
    Yes
    So since they touch 3 times, and a circle is clearly very symmetrical then you know that in one direction it must intersect twice, and in the other direction it must just touch the axis. So what would the radius have to be if it is to touch one of the axis considering the centre? (there's two options here)
    Then considering the two options, which one also means that it intersects the other axis?
    • Thread Starter
    Offline

    2
    ReputationRep:
    Hi!
    When I set x=0 and used b^2-4ac I got an answer of 36, and when I set y=0 I got 25. I don't understand what I do now.

    (Original post by phys981)
    You can do it a different way - I think it's more about quadratic equation solving than circles.

    Where it meets the axis, either x or y must be zero.

    So, put x = 0, this will give you a quadratic in y (1)

    Similarly, putting y = 0 gives you a quadriatic in x. (2)

    Now, if there are only THREE places where it crosses the axes, then there are THREE solutions to those two quadratics which means one of them only has one solution. That means for one of them, the determinant is 0.

    So, I took my two new quadratic equations, found the value of k which gave a determinant of zero.

    That meant I had two possible values of k, one from (1) and one from (2)

    I put the value of k from (1) back into (2) to try to solve for x
    I put the value of (2) back into (1) to try to solve for y

    You don't actually have to find them, just check the determinants again.

    One of those two (not telling you which, try it) gave me a negative determinant which meant the equation would have NO solutions which means the original equation could not have THREE solutions.
    The other one woried. I found the markscheme, checked and it's right.

    Try it - if you can't get it to work out come back and ask again.
    Offline

    10
    ReputationRep:
    Yes - you are right so far.

    I said if k = 25, then the quadratic I got for x only has one solution so the one I got for y must give me two solutions which means the determinant part of that quadratic in y is positive. - check it by substutuing k=25 into it.

    Similarly, if k=36 then the quadratic I got for x must give two solutions so has a positive determinant. Chck it by subs k=36.

    One of them gives you a negative determinant so no solution, therefore the other one must be correct.
    • Thread Starter
    Offline

    2
    ReputationRep:
    So, do I put the value of k into y^2-10y+k and x^2-12x+k?

    (Original post by phys981)
    Yes - you are right so far.

    I said if k = 25, then the quadratic I got for x only has one solution so the one I got for y must give me two solutions which means the determinant part of that quadratic in y is positive. - check it by substutuing k=25 into it.

    Similarly, if k=36 then the quadratic I got for x must give two solutions so has a positive determinant. Chck it by subs k=36.

    One of them gives you a negative determinant so no solution, therefore the other one must be correct.
    Offline

    10
    ReputationRep:
    (Original post by Cameron45)
    So, do I put the value of k into y^2-10y+k and x^2-12x+k?

    Yes, put the 36 into the y² one, then put the 25 into the x² one and you should find that one of them gives a positive determinant, so is correct, and the other gives a negative determinant.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Yeah, the 36 into y^2 gives you a negative, so 25 must be right because it gives you 3 solutions?

    (Original post by phys981)
    Yes, put the 36 into the y² one, then put the 25 into the x² one and you should find that one of them gives a positive determinant, so is correct, and the other gives a negative determinant.
    Offline

    10
    ReputationRep:
    (Original post by Cameron45)
    Yeah, the 36 into y^2 gives you a negative, so 25 must be right because it gives you 3 solutions?
    Yes!!!

    here is the paper it came from https://www.sqa.org.uk/files_ccc/H_M...ellcheckon.pdf Q18 I think

    and here the MS https://www.sqa.org.uk/pastpapers/pa...cs_mi_2015.pdf

    just so you know I'm not kidding.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Same paper i'm working from just now!
    Thank you for your help!

    (Original post by phys981)
    Yes!!!

    here is the paper it came from https://www.sqa.org.uk/files_ccc/H_M...ellcheckon.pdf Q18 I think

    and here the MS https://www.sqa.org.uk/pastpapers/pa...cs_mi_2015.pdf

    just so you know I'm not kidding.
    Offline

    10
    ReputationRep:
    (Original post by Cameron45)
    Same paper i'm working from just now!
    Thank you for your help!
    You're welcome

    But hide that markscheme and do it properly
    • Thread Starter
    Offline

    2
    ReputationRep:
    Will do haha!

    (Original post by phys981)
    You're welcome

    But hide that markscheme and do it properly
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    Lemur14's method is the nicest and fastest for this question (a diagram will help). There's no need to consider discriminants.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 14, 2018
The home of Results and Clearing

3,572

people online now

1,567,000

students helped last year

University open days

  1. Bournemouth University
    Clearing Open Day Undergraduate
    Fri, 17 Aug '18
  2. University of Bolton
    Undergraduate Open Day Undergraduate
    Fri, 17 Aug '18
  3. Bishop Grosseteste University
    All Courses Undergraduate
    Fri, 17 Aug '18
Poll
A-level students - how do you feel about your results?
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

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.