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Given co-ords of only 3 points A, B, C, and value of angle ADC (alpha), how can I find out the coord values x,y of point D ?

Thanks

- Thread starter Emieno
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- #1

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Given co-ords of only 3 points A, B, C, and value of angle ADC (alpha), how can I find out the coord values x,y of point D ?

Thanks

- #2

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oOooo, I forgot to say, the two points at which the two circles intersect are A and D.

- #3

Fermat

Homework Helper

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Let the circle with A,C and D be fixed - in size.

Let the circle through A and B now vary in size such that the point of intersection of the circles is D.

The point D will vary in position, but still be a point somewhere on the circumference.

But by simple geometry, the angle ADC is constant regardlees of the position of D.

In other words, with AB and C fixed in positoin and with angle ADC fixed in size, I have been able to vary the postion of D - ergo no solution

- #4

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Oh sorry, another detail I forgot to say is angle BDA=beta

In sum, A,D,C are on O1, B is on O2, O1 cut O2= {A,D} , BDA=beta, ADC=alpha, A(a1,a2), C(c1,c2),B(b1,b2) and D is what I have to find.

In sum, A,D,C are on O1, B is on O2, O1 cut O2= {A,D} , BDA=beta, ADC=alpha, A(a1,a2), C(c1,c2),B(b1,b2) and D is what I have to find.

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- #5

Fermat

Homework Helper

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There's a little bit of geometry involved at first, then the rest is just a lot of trig.

**Here's the Strategy**

Take one of the circles.

We define a third point on that circle.

Then we set up the eqn for that circle: (x-g)² + (y-h)² = R², where (g,h) are the coords of the centre-point of that circle.

We set up the eqn for the 2nd circle.

Find out where the circles intersect.

A(a1,a2) will be one solution. D(d1,d2) will be the other solution.

**The Geometry bit**

Take any circle. Let P and Q be two points on it. PQ is a chord of that circle. Let R be any other point on the circle. The angle PRQ is the angle subtended by the chord PQ. Now there is a geometry theorem, which I can't for the life of me remember the name of, but it means that wherever R is on the circle, then the angle PRQ is always the same. Here endeth the geometry lesson.

**The applied geometry bit**

Let C1 be the circle, with centre O1, containing the points A, B and D. AB is a chord of that circle. The angle ADB is beta. Now draw a line from A through the centre-point, O1, meeting the circle at the point E, say. Then the line AE is a diameter of the circle, and by the geometry theorem just mentioned, angle AEB, on the chord AB is equal to the angle ADB, also on the chord AB. Also, since AE is a diameter, then angle ABE is a right angle.

**The trig bit**

The rest is now just a lot of trig, to finish off the Strategy.

Can you work out from this, the eqn of the circle defining the circle C1?

Take one of the circles.

We define a third point on that circle.

Then we set up the eqn for that circle: (x-g)² + (y-h)² = R², where (g,h) are the coords of the centre-point of that circle.

We set up the eqn for the 2nd circle.

Find out where the circles intersect.

A(a1,a2) will be one solution. D(d1,d2) will be the other solution.

Take any circle. Let P and Q be two points on it. PQ is a chord of that circle. Let R be any other point on the circle. The angle PRQ is the angle subtended by the chord PQ. Now there is a geometry theorem, which I can't for the life of me remember the name of, but it means that wherever R is on the circle, then the angle PRQ is always the same. Here endeth the geometry lesson.

Let C1 be the circle, with centre O1, containing the points A, B and D. AB is a chord of that circle. The angle ADB is beta. Now draw a line from A through the centre-point, O1, meeting the circle at the point E, say. Then the line AE is a diameter of the circle, and by the geometry theorem just mentioned, angle AEB, on the chord AB is equal to the angle ADB, also on the chord AB. Also, since AE is a diameter, then angle ABE is a right angle.

The rest is now just a lot of trig, to finish off the Strategy.

Can you work out from this, the eqn of the circle defining the circle C1?

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