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NCERT Exemplar · Q7

Q.If z=x+iyz=x+iy, then show that zzˉ+2(z+zˉ)+b=0z\bar{z}+2(z+\bar{z})+b=0, where b∈Rb\in\mathbf{R}, represents a circle.

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The given equation zzˉ+2(z+zˉ)+b=0z\bar{z}+2(z+\bar{z})+b=0 can be rewritten in the form ∣z+2∣2=4−b|z+2|^2 = 4-b, which is the standard equation of a circle with centre at (−2,0)(-2,0) and radius 4−b\sqrt{4-b}, provided b<4b<4.

The key is to recognise that zzˉ=∣z∣2z\bar{z} = |z|^2 and z+zˉ=2Re⁡(z)z+\bar{z} = 2\operatorname{Re}(z). These are the building blocks that let us translate a complex equation into a familiar Cartesian form.

  1. Rewrite using real and imaginary parts.

    Let z=x+iyz = x+iy, so zˉ=x−iy\bar{z} = x-iy. Then:

    • zzˉ=(x+iy)(x−iy)=x2+y2z\bar{z} = (x+iy)(x-iy) = x^2 + y^2
    • z+zˉ=(x+iy)+(x−iy)=2xz+\bar{z} = (x+iy)+(x-iy) = 2x

    Substituting into the given equation:

x2+y2+2(2x)+b=0x^2 + y^2 + 2(2x) + b = 0

x2+y2+4x+b=0x^2 + y^2 + 4x + b = 0

  1. Complete the square in xx. Group the xx terms: x2+4xx^2 + 4x. To complete the square, add and subtract 44:

(x2+4x+4)+y2+b−4=0(x^2 + 4x + 4) + y^2 + b - 4 = 0

(x+2)2+y2=4−b(x+2)^2 + y^2 = 4 - b

  1. Interpret the result. The equation (x+2)2+y2=4−b(x+2)^2 + y^2 = 4-b is precisely the standard form of a circle:
    • Centre: (−2,0)(-2, 0)
    • Radius: 4−b\sqrt{4-b}, which is real only when 4−b>04-b > 0, i.e., b<4b < 4. …

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