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Question: The locus of the center of a circle which touches the circles\[\left| {z - {z_1}} \right| = a\] and ...

The locus of the center of a circle which touches the circles∣z−z1∣=a\left| {z - {z_1}} \right| = a and ∣z−z2∣=b\left| {z - {z_2}} \right| = b externally (z,z1,z2z,{z_1},{z_2} are complex numbers) will be-

Explanation

Solution

Here we will first draw the diagram of the circles which touch the other two circles externally. Then we will see the distance between their centers and from those equations we will find the other equation. Then we will see or check whether the final equation satisfies any of the curve property.

Complete Complete Step by Step Solution:
It is given that the circle touches the other two circle i.e. ∣z−z1∣=a\left| {z - {z_1}} \right| = a and ∣z−z2∣=b\left| {z - {z_2}} \right| = b externally.
So, the center of the circle one is z1{z_1} with radius aa and the center of the other circle is z2{z_2} with radius bb.
Now we will draw these two circles with the third circles which touch them externally. So, we get

Now we know that the distance between the centers of the circle is equal to the sum of their radius. Therefore, we can write it as
⇒∣z−z1∣=a+r\Rightarrow \left| {z - {z_1}} \right| = a + r……………. (1)\left( 1 \right)
Similarly writing the distance equation for other circle, we get
⇒∣z−z2∣=b+r\Rightarrow \left| {z - {z_2}} \right| = b + r……………. (2)\left( 2 \right)
Now from the equation (1)\left( 1 \right) we will find the value of rr . So, we get
r=∣z−z1∣−ar = \left| {z - {z_1}} \right| - a
Substituting the value of rr in the equation (2)\left( 2 \right), we get
∣z−z2∣=b+r=b+∣z−z1∣−a\left| {z - {z_2}} \right| = b + r = b + \left| {z - {z_1}} \right| - a
Now subtracting ∣z−z1∣\left| {z - {z_1}} \right| from both the sides, we get
⇒∣z−z2∣−∣z−z1∣=b−a\Rightarrow \left| {z - {z_2}} \right| - \left| {z - {z_1}} \right| = b - a
Now, from equation (2)\left( 2 \right), we get
r=∣z−z2∣−br = \left| {z - {z_2}} \right| - b
Substituting this value in equation (1)\left( 1 \right), we get
∣z−z1∣=a+∣z−z2∣−b\left| {z - {z_1}} \right| = a + \left| {z - {z_2}} \right| - b
Now subtracting ∣z−z1∣\left| {z - {z_1}} \right| from both the sides, we get
⇒∣z−z2∣−∣z−z1∣=b−a\Rightarrow \left| {z - {z_2}} \right| - \left| {z - {z_1}} \right| = b - a
Now from both the equations, we got the same result.
This means that the difference between the distances of the center of the circles is constant.
As we know that the distance between the foci of the hyperbola remains constant i.e. ∣PS−PS′∣=2a\left| {PS - PS'} \right| = 2a.
Then by comparing this with the obtained equation, we can say that the locus of the center of a circle which touches the circles∣z−z1∣=a\left| {z - {z_1}} \right| = a and ∣z−z2∣=b\left| {z - {z_2}} \right| = b externally will be a Hyperbola.

Hence, the locus of the center of a circle which touches the circles externally will be a Hyperbola.

Note:

  1. Here we should know that the distance between the foci of the hyperbola always remains constant i.e. ∣PS−PS′∣=2a\left| {PS - PS'} \right| = 2a also the distance between the centers of the two external touching circles is equal to the sum of their radius.
  2. While solving this question we have to make an equation that satisfies the property of the curve then we can say that the locus of the center of a circle is of that type of curve.