Maths-

General

Easy

Question

# A (1, 2) B (-1, 2) are two fixed points. The locus of P such that PA = n. PB, where n ¹ 1 is a constant, is1

- not a circle
- straight line
- circle
- parabola

Hint:

### evaluate the expression given and solve the resultant expression to get the required locus.

## The correct answer is: circle

### circle

Let p be (x,y)

PA= √(x-1)^{2}+(y-2)^{2}

PB = √(x+1)^{2}+(y-2)^{2}

PA=nPB

=> √(x-1)^{2}+(y-2)^{2 }=n (√(x+1)^{2}+(y-2)^{2})

On squaring both sides, we get

(x-1)^{2}+(y-2)^{2}=n^{2}((x+1)^{2}+(y-2)^{2})

On further simplification, we get

X^{2}(1-n^{2})+y^{2}(1-n^{2 })+2x(-n^{2}-1)+2y(2n^{2 }-2)+(1+n^{2}+4+4)=0

This is of the form of locus of a circle.

**x2 + y2 + 2gx + 2fy + c = 0 is the general equation of a circle.**

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