Finding Intersection of Tangents on a Circle

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SUMMARY

The discussion focuses on finding the intersection point of tangents to a circle defined by the equation |z|=1 at points z₁ and z₂. The participants clarify that while z₁ and z₂ lie on the circle centered at (0,0), the intersection point z₃ does not lie on this circle. The correct approach involves deriving the equations of the tangent lines and solving for their intersection point, leading to the expression for z₃ in terms of z₁ and z₂. The final derived formula for z₃ is z₃ = (2(z₂ - z₁ - z̅₁)) / (z̅₁z₂ - z₁z̅₂), although it was noted that this result may contain errors.

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utkarshakash
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Homework Statement


Two points represented by z_{1}, z_{2} lie on circle |z|=1, the tangents to the circle at these points meet at the point represented by

Homework Equations





The Attempt at a Solution


Let the tangents meet at the point z_{3}. The centre of the given circle is (0,0).
∴ z_{1}, z_{2}, z_{3} and 0 are concyclic. Now what to do next?
 
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utkarshakash said:

Homework Statement


Two points represented by z_{1}, z_{2} lie on circle |z|=1, the tangents to the circle at these points meet at the point represented by

Homework Equations





The Attempt at a Solution


Let the tangents meet at the point z_{3}. The centre of the given circle is (0,0).
∴ z_{1}, z_{2}, z_{3} and 0 are concyclic. Now what to do next?

It is not true. z1 and z2 are on the circle which centre is (0,0) z3 can not lie on the circle.

Find the equations of the tangent line and the point where they cross each other.

ehild
 
ehild said:
It is not true. z1 and z2 are on the circle which centre is (0,0) z3 can not lie on the circle.

Find the equations of the tangent line and the point where they cross each other.

ehild

I'm not saying that z_{3} will lie on the circle with centre (0,0), Instead the 4 points which I mentioned will lie on some other circle and the centre of that circle will be the mid-point of line joining z_{3} and 0.
 
Sorry, I misunderstood you. You are right, z1,z2, (0,0)and z3 are on a circle. Find z3 in terms of z1 and z2. One possible way is to write up the equations of the tangent lines and find the intersection. Do you know the equation of a straight line in term of its normal? ehild
 
Last edited:
ehild said:
Sorry, I misunderstood you. You are right, z1,z2, (0,0)and z3 are on a circle. Find z3 in terms of z1 and z2. One possible way is to write up the equations of the tangent lines and find the intersection. Do you know the equation of a straight line in term of its normal?


ehild

I know how to write equation of tangents in terms of x and y but not in the form of z. But I know the equation of a complex line.
 
A complex number has x and y components like a vector in a plane. Write up the equations of the tangents in terms of x1,y1 and x2,y2. The coordinates of the point of intersection are x3,y3. You can convert it to the complex number z3=x3+iy3.

ehild
 
ehild said:
A complex number has x and y components like a vector in a plane. Write up the equations of the tangents in terms of x1,y1 and x2,y2. The coordinates of the point of intersection are x3,y3. You can convert it to the complex number z3=x3+iy3.

ehild

But I have to give the answer in terms of z1 and z2 and converting the answer to the required form will be calculative. Any other methods or should I go with this?
 
Do it as you can and we will see.

ehild
 
ehild said:
Do it as you can and we will see.

ehild

After writing the equation in terms of x and y and then simultaneously solving the two lines I get


\LARGE z_{3}= \frac{y_{2}-y_{1}}{x_{1}y_{2}-x_{2}y_{1}} + i \frac{x_{1}-x_{2}}{x_{1}y_{2}-x_{2}y_{1}}



Now how do I represent this in terms of z1 and z2?
 
  • #10
z=x+iy. x=Re(z) or (z+z*)/2; y=Im(z) or y=(z-z*)/2i. (* means complex conjugate)

ehild
 
  • #11
ehild said:
z=x+iy. x=Re(z) or (z+z*)/2; y=Im(z) or y=(z-z*)/2i. (* means complex conjugate)

ehild

After simplification I get

\LARGE z_{3}= \frac{2(z_{2}-z_{1}- \overline{z_{1}})}{\overline{z_{1}}z_{2}-z_{1} \overline{z_{2}}}

But this is not the correct answer.
 
  • #12
utkarshakash said:
After simplification I get

\LARGE z_{3}= \frac{2(z_{2}-z_{1}- \overline{z_{1}})}{\overline{z_{1}}z_{2}-z_{1} \overline{z_{2}}}

But this is not the correct answer.

Your result is not correct. Check the signs in your derivation.

ehild
 

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