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    C is a circle with the equation x^2 + y^2 - 2x - 10y + 21.

    The equation of this circle is (x-1)^2 + (y-5)^2 = 5

    Given that the line joining P(3,6) and Q(q,4) is a diameter of C, show that q = -1.

    So far, I've tried using Pythagoras Theorem to find q, since I know that the length of the diameter is √5.

    Like so:

    (3-q)^2 + (6-4)^2 = 2√5.

    But this brings me to a quadratic, and two incorrect solutions.

    What am I missing here?
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    (Original post by MathFora)
    C is a circle with the equation x^2 + y^2 - 2x - 10y + 21.

    The equation of this circle is (x-1)^2 + (y-5)^2 = 5

    Given that the line joining P(3,6) and Q(q,4) is a diameter of C, show that q = -1.

    So far, I've tried using Pythagoras Theorem to find q, since I know that the length of the diameter is √5.

    Like so:

    (3-q)^2 + (6-4)^2 = 2√5.

    But this brings me to a quadratic, and two incorrect solutions.

    What am I missing here?
    Two things:

    With the method you're using, your equation should be (3-q)^2+(6-4)^2=(2\sqrt{5})^2

    However, since PQ is a diameter, P is a far to the right and up from the centre, as Q will be to the left and down. Similar triangles if you want to drawn in the detail. No quadratic required.
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    What you're missing is that the (x,y) displacement from P to the centre of the circle will be the same as the (x,y) displacement from the centre of the circle to Q (as PQ is a diameter).

    Your rearranged circle equation gives you the coordinates of the centre.

    (Overlapped with Ghostwalker's response).
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    (Original post by ghostwalker)
    Two things:

    With the method you're using, your equation should be (3-q)^2+(6-4)^2=(2\sqrt{5})^2

    However, since PQ is a diameter, P is a far to the right and up from the centre, as Q will be to the left and down. Similar triangles if you want to drawn in the detail. No quadratic required.
    Thanks!

    I've solved the question using a quadratic because that's the method I started doing it with.

    Two more questions : Is q = -1 for one of the solutions of the quadratic enough to prove that q = -1?

    Also, what is the other solution of the quadratic, 7?
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    (Original post by MathFora)
    Thanks!

    I've solved the question using a quadratic because that's the method I started doing it with.

    Two more questions : Is q = -1 for one of the solutions of the quadratic enough to prove that q = -1?

    Also, what is the other solution of the quadratic, 7?
    Well you have two solutions. So, you'd need to check which one lies on the circle to confirm q= -1.

    The other solution arises, because there are two points on the line y=4, that are a distance 2root(5) from P.
 
 
 
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