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Im(iz-2/z-1)+...

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Im(iz-2/z-1)+1=0,z belongs to C,z not equals to i represents a part of circle having radius equal to:

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Q. The equation

I m (\frac{i z - 2}{z - i}) + 1 = 0, z ϵ C, z \neq i

represents a part of a circle having radius equal to.

\frac{| z - 2 |}{| z - 3 |} = 2

represents a circle, then its radius is equal to

Q. Let the lines

(2 - i) z = (2 + i) ¯ ¯ ¯ z

(2 + i) z + (i - 2) ¯ ¯ ¯ z - 4 i = 0,

(

i^{2} = - 1

)

be normal to a circle

C

i z + ¯ z + 1 + i = 0

is tangent to this circle

C

, then its radius is :

| z | = 2

and locus of 5z - 1 is a circle having radius `a' and

z_{1}^{2} + z_{2}^{2} - 2 z_{1} z_{2} cos θ = 0

| z_{1} | : | z_{2} |

Q. Let the lines

(2 - i) z = (2 + i) ¯ ¯ ¯ z

(2 + i) z + (i - 2) ¯ ¯ ¯ z - 4 i = 0,

(

i^{2} = - 1

)

be normal to a circle

C

i z + ¯ z + 1 + i = 0

is tangent to this circle

C

, then its radius is :

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