A(-a,0) and B(a,0) are basically two points on the x-axis which are equidistant from the origin (0,0)

Let (x,y) be the coordinates of P.

PA = 2PB

Using linear distance formula between two points (which is based on Pythagoras' Theorem) ,


√[(x - (-a))² + (y - 0)²] = 2√[(x - a)² + (y - 0)²]

√[(x + a)² + y²] = 2√[(x - a)² + y²]

Square both sides,


(x + a)² + y² = 4[(x - a)² + y²]

(x + a)² + y² = 4(x - a)² + 4y²

x² + 2ax + a² + y² = 4x² - 8ax + 4a² + 4y²

3x² + 3y² - 10ax + 3a² = 0

x² + y² - 10a/3 x + a² = 0


We realise that we have just obtained the general form of the equation of a circle,

x² + y² + 2gx + 2fy + c = 0,

where g = -5a/3 , c = a², f = 0


The centre has coordinates (-g,-f)

So the centre is (5a/3 ,0)

c = g² + f² - r² , where r is the radius of the circle.

So r² = g² + f² - c

= (-5a/3)² + 0² - a²
= 25a²/9 - a²
= 16a²/9

r = √(16a²/9)

r = √((4a/3)²)

r = |4a/3|


(Since we do not know if a is positive or negative, put the modulus sign to get the absolute value/magnitude which is positive. Radius has to be positive)


Alternatively you could write r = 4a/3 for a > 0 and r = -4a/3 for a < 0


So since P satisfies PA = 2PB and P also satisfies the circle equation, all possible points of P are found on a circle which means the path traced is essentially that circle itself.

Alternatively,

From earlier working, rewrite as :

x² - 10a/3 x + y² = -a²


Completing the square,

(x² - 2(5a/3)x + (5a/3)²) - (5a/3)² + (y - 0)² = -a²

(x - 5a/3)² + (y - 0)² = -a² + (5a/3)²

(x - 5a/3)² + (y - 0)² = -a² + 25a²/9

(x - 5a/3)² + (y - 0)² = 16a²/9

(x - 5a/3)² + (y - 0)² = (4a/3)²


This is in the standard form of the equation of a circle

(x - h)² + (y - k)² = r²

Where the centre has coordinates (h,k) and radius is r

So centre is (5a/3, 0) and r² = (4a/3)²

radius = √((4a/3)²) = |4a/3|


Or

radius = 4a/3 for a > 0 and -4a/3 for a < 0