Skip to content
Exercise 2.6 · Q5

Q.Obtain the Cartesian equation for the locus of z=x+iyz=x+iy in each of the following cases:

(i) ∣z−4∣=16|z-4|=16
(ii) ∣z−4∣2−∣z−1∣2=16|z-4|^2-|z-1|^2=16.
Puducherry TnboardTextbookSubjectiveImportance★★★★★
28% · 34/122 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

  1. is a circle equation in disguise; substitute z=x+iyz=x+iy and expand directly.
  2. is a difference of two squared-modulus expressions, which telescopes into a linear equation in xx once expanded via the difference-of-squares identity.

Step 1. (i) Substitute z=x+iyz=x+iy into ∣z−4∣=16|z-4|=16. z−4=(x−4)+iyz-4=(x-4)+iy, so ∣z−4∣=(x−4)2+y2=16|z-4|=\sqrt{(x-4)^2+y^2}=16.

Step 2. (i) Square both sides and expand.

(x−4)2+y2=256 ⇒ x2−8x+16+y2=256 ⇒ x2+y2−8x−240=0.(x-4)^2+y^2=256\ \Rightarrow\ x^2-8x+16+y^2=256\ \Rightarrow\ x^2+y^2-8x-240=0.

(This is the circle of radius 1616 centred at (4,0)(4,0), written in Cartesian form.)

Step 3. (ii) Write ∣z−4∣2|z-4|^2 and ∣z−1∣2|z-1|^2 in terms of x,yx,y. ∣z−4∣2=(x−4)2+y2|z-4|^2=(x-4)^2+y^2 and ∣z−1∣2=(x−1)2+y2|z-1|^2=(x-1)^2+y^2.

Step 4. (ii) Subtract — the y2y^2 terms cancel.

∣z−4∣2−∣z−1∣2=(x−4)2−(x−1)2.|z-4|^2-|z-1|^2=(x-4)^2-(x-1)^2. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.