__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 (*x*) is:_{1}, y_{1}**2.**The equation of the locus of a moving point

*P*(

*x, y*) which is

always at a constant distance from two fixed points (

*x*) and_{1}, y_{1} (

*x*) with a ratio is:_{1}, y_{1}**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*of the straight line*

**perpendicular bisector***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*–*x*_{1})^{2}+ (*y*–*y*_{1})^{2}=*r*^{2}(

*x*– 2)^{2}+ (*y*– 4)^{2}= 5^{2}*x*

^{2}– 4

*x*+ 4 +

*y*

^{2}– 8

*y*+ 16 = 25

*x*

^{2}+

*y*

^{2}– 4

*x*– 8

*y*– 5 = 0

**Example 2**

Find the equation of the locus of a moving point

$\begin{array}{l}\text{Given}PA=PB\\ \sqrt{{\left(x-\left(-2\right)\right)}^{2}+{\left(y-3\right)}^{2}}=\sqrt{{\left(x-4\right)}^{2}+{\left(y-\left(-1\right)\right)}^{2}}\\ \text{Squarebothsidestoeliminatethesquareroots}\text{.}\\ {\left(x+2\right)}^{2}+{\left(y-3\right)}^{2}={\left(x-4\right)}^{2}+{\left(y+1\right)}^{2}\\ {x}^{2}+2x+4+{y}^{2}-6y+9={x}^{2}-8x+16+{y}^{2}+2y+1\\ 10x-8y-4=0\\ \text{Hence,theequationofthelocusofpoint}P\text{is}\\ 10x-8y-4=0\end{array}$
*P*(*x, y*) which is always equidistant from points*A*(-2, 3) and*B*(4, -1).*Solution:***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:*