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6.6 Equation of a Locus

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6.6 Equation of a Locus
1. The equation of the locus of a moving point P(x, y) which is always at a constant distance (r) from a fixed point (x1, y1) is:


2. The equation of the locus of a moving point P(x, y) which is always at a constant distance from two fixed points (x1, y1) and  (x1, y1) with a ratio is:



3. The equation of the locus of a moving point P (x, y) which is always equidistant from two fixed points A and B is the perpendicular bisector of the straight line AB.


Example 1
Find the equation of the locus of a moving point P(x, y) which is always at a distance of 5 units from a fixed point Q (2, 4).

Solution:
(x – x1)2+ (y – y1)2 = r2
(x – 2)2 + (y – 4)2 = 52
x2 – 4x + 4 + y2 – 8y + 16 = 25
x2 + y2– 4x – 8y – 5 = 0



Example 2
Find the equation of the locus of a moving point P(x, y) which is always equidistant from points A (-2, 3) and B (4, -1).

Solution:
Given PA=PB√(x−(−2))2+(y−3)2=√(x−4)2+(y−(−1))2Square both sides to eliminate the square roots.(x+2)2+(y−3)2=(x−4)2+(y+1)2x2+2x+4+y2−6y+9=x2−8x+16+y2+2y+110x−8y−4=0Hence, the equation of the locus of point P is10x−8y−4=0


Example 3
A (2, 0) and B (0, -2) are two fixed points. Point P moves with a ratio so that AP: PB = 1: 2.  Find the equation of the locus of point P.

Solution:
AP:PB=1:2APPB=122AP=PB2√(x−2)2+(y−0)2=√(x−0)2+(y−(−2))2Square both sides to eliminate the square roots.4[(x−2)2+y2]=x2+(y+2)24(x2−4x+4+y2)=x2+y2+4y+44x2−16x+16+4y2=x2+y2+4y+43x2+3y2−16x−4y+12=0Hence, the equation of the locus of point P is3x2+3y2−16x−4y+12=0