Results 1 to 6 of 6

Math Help - circle

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    56

    circle

    Let C be the circle in R having the point (h, k) and (0, 1) as diameter. Prove that this circle intersects the x-axis if and only if h-4k≥0 and in this case the two intercepts are the roots of the equation x-hx+k=0.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    First, it doesn't make sense. I am guessing the diameter is the line segment between (h,k) and (0,1)

    Second, isn't this just an algebra problem?

    Center of Circle

    \left(\frac{h}{2},\frac{k+1}{2}\right)

    Radius of Circle

    \sqrt{(h-0)^{2}+(k-1)^{2}}

    Where's the tricky part?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member grgrsanjay's Avatar
    Joined
    May 2010
    From
    chennai,tamil nadu
    Posts
    143
    Thanks
    1

    Red face confused

    i cant even understand the question is that the question says that find the radius/
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,598
    Thanks
    1421
    Quote Originally Posted by apple2009 View Post
    Let C be the circle in R having the point (h, k) and (0, 1) as diameter.
    As TKHunny says, two "points" do not form a diameter- you mean the line segment having those points as endpoints.

    Prove that this circle intersects the x-axis if and only if h-4k≥0 and in this case the two intercepts are the roots of the equation x-hx+k=0.
    The center of a circle is the midpoint of any diameter so this circle has \left(\frac{h+0}{2}, \frac{k+1}{2}\right)= \left(\frac{h}{2}, \frac{k+1}{2}\right) as center.

    The radius of a circle is the distance from the center to any point on the circumference. Since \left(\frac{h}{2}, \frac{k+1}{2}\right) is the center and (0, 0) is on the circumference, the radius is \sqrt{\frac{h^2}{4}+ \frac{(k+1)^2}{4}}

    The equation of a circle with center at (a, b) and radius r is (x- a)^2+ (y- b)^2= r^2 so the equation of this circle is \left(x- \frac{h}{2}\right)^2+ \left(y- \frac{k+1}{2}\right)^2= \frac{h^2+ (k+1)^2}{4}.

    The first thing we can do is multiply both sides of that by 4, leaving (2x- h)^2+ (2y- (k+1))^2= h^2+ (k+1)^2.

    The circle "intersects the x-axis" where y= 0. Put y= 0 into that equation and solve for x. That will be a quadratic equation so it will have two distinct solutions if and only its discrimant is 0. (The "discriminant" of the quadrtic equation, ax^2+ bx+ c= 0 is b^2- 4ac.)
    Last edited by HallsofIvy; May 6th 2010 at 07:59 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member grgrsanjay's Avatar
    Joined
    May 2010
    From
    chennai,tamil nadu
    Posts
    143
    Thanks
    1

    Thumbs up thanks

    thank you very much
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Oct 2009
    Posts
    459
    There is a circle equation formed from 2 end points of a diameter but it does not seem to be regularly taught.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Circle, tangent line, and a point not on the circle
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: March 31st 2011, 01:40 PM
  2. Replies: 6
    Last Post: July 8th 2010, 05:39 PM
  3. Replies: 7
    Last Post: March 15th 2010, 04:10 PM
  4. Replies: 2
    Last Post: February 6th 2010, 08:31 AM
  5. Replies: 0
    Last Post: October 12th 2008, 04:31 PM

Search Tags


/mathhelpforum @mathhelpforum