Question Help?: Locus (1 Viewer)

[cricket_freak]

Can someone pl help me with question: Find the equation of the locus of a point that moves so that it is always 2 units from the point (5, -2).

 

[Riviet]

This is exactly the same question as the other that you posted up before, just different numbers, that's all. Simply do the same method as before:

the equation of this locus takes the form of the equation of the circle.
i.e (x-h)² + (y-k)² = r², where h and k are the x and y values of the centre of the circle and r is the length of the radius. since the distance from the point (5,-2) is always 2 units, then the radius is 2 and the h and k values of the equation are 5 and -2 respectively

Substituting into the general equation of a circle,
(x-5)² + (y-[-2])² = 2²
(x-5)² + (y+2)² = 4

.'. The locus is a circle with centre (5,-2) and radius 2 units.

 

[Ioup]

(x-5)squared+(y+2)squared=4

I cant remember if thats right or not.
Did locus far back some time.
You basically just have to remember that the locus will carry out a path of a circle.

 

[pLuvia]

If you don't realise that the locus is a circle use the distant formula, treating one of the points as P(x,y)

√(x-5)²+(y+2)² = 2

Square both sides

(x-5)²+(y+2)² = 2²
This is the general equation of a circle hence:

The locus is a circle with centre (5,-2) and radius 2

 

[vizman]

substitute into circle formula (x-a)^2 + (y-b)^2 = r^2