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Question: Inverse of a point a with respect to the circle \(|z - c| = R\) (a and c are complex numbers, centre...

Inverse of a point a with respect to the circle ∣z−c∣=R|z - c| = R (a and c are complex numbers, centre C and radius R) is the point c+R2aˉ−cˉc + \frac{R^{2}}{\bar{a} - \bar{c}}

A

c+R2aˉ−cˉc + \frac{R^{2}}{\bar{a} - \bar{c}}

B

c−R2aˉ−cˉc - \frac{R^{2}}{\bar{a} - \bar{c}}

C

c+Rcˉ−aˉc + \frac{R}{\bar{c} - \bar{a}}

D

None of these

Answer

c+R2aˉ−cˉc + \frac{R^{2}}{\bar{a} - \bar{c}}

Explanation

Solution

Sol. Let a' be the inverse point of a with respect to the circle ∣z−c∣=R,|z - c| = R, then by definition the points c, a, a' are collinear.

We have, a⥂rg(a′−c)=a⥂rg(a−c)=−a⥂rg(aˉ−cˉ)a ⥂ rg(a' - c) = a ⥂ rg(a - c) = - a ⥂ rg(\bar{a} - \bar{c}) (∵a⥂rgzˉ=−a⥂rgz)(\because a ⥂ rg\bar{z} = - a ⥂ rgz)

⇒ a⥂rg(a′−c)+a⥂rg(aˉ−cˉ)=0a ⥂ rg(a' - c) + a ⥂ rg(\bar{a} - \bar{c}) = 0 ⇒ a⥂rg{(a′−c)(aˉ−cˉ)}=0a ⥂ rg\{(a' - c)(\bar{a} - \bar{c})\} = 0

∴ (a′−c)(aˉ−cˉ)(a' - c)(\bar{a} - \bar{c}) is purely real and positive.

By definition ∣a′−c∣∣a−c∣=R2|a' - c||a - c| = R^{2}

(∵CP.C⥂Q=r2)(\because CP.C ⥂ Q = r^{2})

⇒ ∣a′−c∣∣aˉ−cˉ∣=R2|a' - c||\bar{a} - \bar{c}| = R^{2} (∵∣z∣=∣zˉ∣)(\because|z| = |\bar{z}|)

⇒ ∣(a′−c)(aˉ−cˉ)∣=R2|(a' - c)(\bar{a} - \bar{c})| = R^{2} ⇒ (a′−c)(aˉ−cˉ)=R2(a' - c)(\bar{a} - \bar{c}) = R^{2} {∵(a′−c)(aˉ−cˉ)\{\because(a' - c)(\bar{a} - \bar{c}) is purely real and positive}

⇒ a′−c=R2aˉ−cˉa' - c = \frac{R^{2}}{\bar{a} - \bar{c}}. Therefore, the inverse point a' of a point a, a′=c+R2aˉ−cˉ.a' = c + \frac{R^{2}}{\bar{a} - \bar{c}}.