Finding Intersection of Tangents on a Circle

[utkarshakash]

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?

 

[ehild]

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

 

[utkarshakash]

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.

 

[ehild]

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

 

[utkarshakash]

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.

 

[ehild]

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

 

[utkarshakash]

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?

 

[ehild]

Do it as you can and we will see.

ehild

 

[utkarshakash]

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?

 

[ehild]

z=x+iy. x=Re(z) or (z+z*)/2; y=Im(z) or y=(z-z*)/2i. (* means complex conjugate)

ehild

 

[utkarshakash]

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.

 

[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.

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

ehild