**Locus – Problems**

**1)** Find the equation of locus of a point which is a distance 5 from A (4, -3)

**Solution:** Let p (x, y) be a point in the locus

Given,

A (4, -3)

CP = 5

(x – 4)² + (y + 3)² = 25

x² – 8x + 16 + y² + 6y + 9 – 25 = 0

x² + y² – 8x + 6y = 0

**2)** Find the equation of locus a point which is equidistance from the point A (-3, 2) and B (0, 4)

**Solution: **Given points are A (-3, 2), B (0, 4)

let P (x, y) be any point in the locus

PB = PA

PB² = PA²

(x + 3)² + (y – 2)² = (x – 0)² + (y – 4)²

x² + 6x + 9 + y² – 4y + 4 = x² + y² – 8y + 16

6x + 4y = 3 is the equation of the locus

**3)** Find the equation of locus of a point which is equidistance from the coordinate axes

**Solution:** Let P (x, y) be any point in the locus

Let PM-perpendicular distance of P from X-axis = |x|

Let PN-perpendicular distance of P from Y-axis = |y|

Given PM = PN

x² – y² = 0.