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Circle Question watch

1. 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?
2. (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

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.
3. 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).
4. (Original post by ghostwalker)
Two things:

With the method you're using, your equation should be

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?
5. (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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Updated: October 5, 2017
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