Question

find the equation of locus of point p the distance of p from origin is twice the distance of p from A{1,2}​

Answer 1

Let us consider that point P is (x, y).

a/c to question,

distance between P and origin = 2 × distance between P and A(1, 2)

we know distance formula

d = $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

so, distance between P and origin = √{(x - 0)² + (y - 0)²} = √(x² + y²)

Distance between P and A = √{(x - 1)² + (y - 2)²}

now, √(x² + y²) = 2 × √{(x - 1)² + (y - 2)²}

squaring both sides,

x² + y² = 4[(x - 1)² + (y - 2)²]

or, x² + y² = 4[x² - 2x + 1 + y² - 4y + 4]

or, x² + y² = 4x² + 4y² - 8x - 16y + 20

or, 3x² + 3y² - 8x - 16y + 20 = 0

or, x² + y² - (8/3)x - (16/3)y + (20/3) = 0 This is the locus of point P. here it is clear that P is equation of circle. so, locus of point is a equation of circle .

Answer 2

Answer:

Step-by-step explanation:

Let N(x,y) be a point on the locus.

Distance between P and origin = 2 × distance between P and A(1, 2)

d = √ ( x-a)² + (y-b)²

Therefore, distance between P and origin = √{(x - 0)² + (y - 0)²} = √(x² + y²)

Distance between P and A = √{(x - 1)² + (y - 2)²}

= √(x² + y²) = 2 × √{(x - 1)² + (y - 2)²}

Squaring both the sides,

= x² + y² = 4[(x - 1)² + (y - 2)²]

= x² + y² = 4[x² - 2x + 1 + y² - 4y + 4]

= x² + y² = 4x² + 4y² - 8x - 16y + 20

= 3x² + 3y² - 8x - 16y + 20 = 0

= x² + y² - (8/3)x - (16/3)y + (20/3) = 0

Therefore, the equation of locus of point p is  x² + y² - (8/3)x - (16/3)y + (20/3).