Question

Two circles $z\bar z + z{\bar a_1} + z{\bar a_1} + {b_1} = 0,b \in R$ and $z\bar z + z{\bar a_2} + z{\bar a_2} + {b_2} = 0,{b_2} \in R$intersects orthogonally, then prove that $\operatorname{Re} ({a_1}{\bar a_2}) = {b_1} + {b_2}$.

Answer 1

Two circles $z\bar z + z{\bar a_1} + z{\bar a_1} + {b_1} = 0,b \in R$ and $z\bar z + z{\bar a_2} + z{\bar a_2} + {b_2} = 0,{b_2} \in R$intersects orthogonally, then prove that $\operatorname{Re} ({a_1}{\bar a_2}) = {b_1} + {b_2}$.

Complete step by step answer:
As per the given question we know that the general centre of the circle is ${z_0} = - a$ and the radius is $r = \sqrt {a\bar a - b}$.
So in the first circle $z\bar z + z{\bar a_1} + z{\bar a_1} + {b_1} = 0$, the centre of the circle is $- {a_1}$ and the radius is $\sqrt {{a_1}{{\bar a}_2} - {b_2}}$.
Similarly in the second circle the centre of the circle is $- {a_2}$ and the radius is $\sqrt {{a_2}{{\bar a}_2} - {b_2}}$.
We know that these circles will intersect orthogonally if the sum of the squares of radii is equal to the square of the distance between their centres. Therefore by substituting the values we have: ${\left| {{a_1} - {a_2}} \right|^2} = {a_1}{\bar a_1} - {b_1} + {a_2}{\bar a_2} - {b_2}$.
By squaring and breaking the brackets we have: $\Rightarrow {a_1}{\bar a_1} + {a_2}{\bar a_2} - {a_1}{\bar a_2} - {a_1}{\bar a_2} = {a_1}{\bar a_1} + {a_1}{\bar a_2} - {b_1} - {b_1}$
On further solving, ${a_1}{\bar a_2} + {\bar a_1}{a_2} = {b_1} + {b_2}$. We can write it as ${a_1}{\bar a_2} + \overline {{a_1}{{\bar a}_2}} = {b_1} + {b_2}$.
So we have $2\operatorname{Re} \left( {{a_1}{{\bar a}_2}} \right) = {b_1} + {b_2}$.
Hence it is proved that $2\operatorname{Re} \left( {{a_1}{{\bar a}_2}} \right) = {b_1} + {b_2}$.

Note: Before solving this kind of question we should be totally aware of the orthogonal circles, their general equation and their radius and centres. We should know the formula above used in the basis of the theorem of the circle that if the sum of the squares of radii is equal to the square of the distance between their centres.