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Math Help - problems about further coordinate geometry(circle)

  1. #1
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    problems about further coordinate geometry(circle)

    there are two questions.

    1)
    A diameter of a circle has end points at A(-5,0) and B(9,0).
    The equation of the circle is (x^2) + (y^2) - 2x + 2y -3 = 0 .
    If C(2,k) lies on the circle, find the value of k and hence prove that triangle ABC is an isosceles right-angled triangle.

    my problems are the one highlighted in red and
    my answer for k is 7 or -7. so, how can I reject k= -7 ??
    i get the equation of the circle by myself, so it might be wrong, if i'm careless, that is.


    2)
    Given two points P(1,4) , Q(-1,-2). PQ is the diameter.
    Show that R(-3,2) lies on the circle and hence prove that angle PRQ is 90 degrees.


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  2. #2
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    Quote Originally Posted by wintersoltice View Post
    there are two questions.

    1)
    A diameter of a circle has end points at A(-5,0) and B(9,0).
    The equation of the circle is (x^2) + (y^2) - 2x + 2y -3 = 0 .
    If C(2,k) lies on the circle, find the value of k and hence prove that triangle ABC is an isosceles right-angled triangle.

    [COLOR=Navy]my problems are the one highlighted in red and
    my answer for k is 7 or -7. so, how can I reject k= -7 ??
    i get the equation of the circle by myself, so it might be wrong, if i'm careless, that is.

    ...
    Since A and B are the endpoints of a diameter the center of the circle is the midpoint of \overline{AB} . M\left(\frac{-5+9}{2}~,~\frac{0+0}{2}\right)=(2,0)

    The length of \overline{AB} is 14 thus the radius r = 7

    Therefore the equation of the circle is: (x-2)^2+y^2=7^2

    There are indeed 2 points C that means you have C_1(2,7) or C_2(2,-7)

    These two points C form a diameter perpendicular to \overline{AB}. That means AC_1BC_2 is an inscribed square which consist of 2 isosceles right triangles.
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  3. #3
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    Quote Originally Posted by wintersoltice View Post
    ...

    2)
    Given two points P(1,4) , Q(-1,-2). PQ is the diameter.
    Show that R(-3,2) lies on the circle and hence prove that angle PRQ is 90 degrees.


    1. Calculate the coordinates of the center.
    2. Calculate the length of PQ.
    3. Calculate the length of the radius.
    4. Calculate the equation of the circle.
    5. Plug in the coordinates of R and check if the equation is true.
    6. Check if (\overline{PR})^2 + (\overline{RQ})^2 = (\overline{PQ})^2
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