f.5 maths Locus

1)
Given two points P(-4, 2)and Q(-1, -6),find the equation of the locus of a point C such that
a) PC = QC,b) PC : QC = 2 : 3

Solution :
a)
Let (x, y) be the coordinates of point C.

PC = QC
√[(x + 4)² + (y - 2)²] =√[(x + 1)² + (y + 6)²]
(x + 4)² + (y - 2)² = (x+ 1)² + (y + 6)²
x² + 8x + 16 + y² -4y + 4 = x² + 2x + 1 + y² +12y + 36
Locus of point C : 6x - 16y - 17 = 0

b)
Let (x, y) be the coordinates of point C.
PC : QC = 2 : 3
√[(x + 4)² + (y- 2)²] : √[(x + 1)² + (y+ 6)²] = 2 : 3
[(x + 4)² + (y - 2)²] :[(x + 1)² + (y + 6)²] =4 : 9
(x² + 8x + 16 + y² -4y + 4) : (x² + 2x + 1 + y² +12y + 36) = 4 : 9
9(x² + y² +8x - 4y + 20) = 4(x² + y² +2x + 12y + 37)
9x² + 9y² + 72x- 36y + 180 = 4x² + 4y² + 8x+ 48y + 148
Locus of point C : 5x² + 5y² + 64x- 84y + 32 = 0


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2)
For all the circles passing through two pointsA(3, 7)and B(6, -5),find the equation of the locus of their centres.

Solution :
Let (x, y) be the coordinates of the centre.

The centre is equidistant from A and B.
√[(x - 3)² + (y - 7)²] =√[(x - 6)² + (y + 5)²]
(x - 3)² + (y - 7)² = (x- 6)² + (y + 5)²
x² - 6x + 9 + y² - 14y+ 49 = x² - 12x + 36 + y² + 10y+ 25
6x - 24y - 3 = 0
Locus of their centres : 2x - 8y -1 =0


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3)
Given two points A(5,3) and B(7,7), find theequation of the locus of a point P such that AP² + BP² = AB².

Solution :
Let (x, y) be the coordinates of point P.
AP² + BP² = AB²
{√[(x - 5)² + (y - 3)²]}² + {√[(x - 7)² + (y- 7)²]}² = {√[(5 - 7)² + (3- 7)²]}²
(x - 5)² + (y- 3)² + (x - 7)² + (y- 7)² = (5 -7)² + (3 - 7)²
x² - 10x + 25 + y² - 6y+ 9 + x² - 14x + 49 + y² - 14y+ 49 = 4 + 16
2x² + 2y² - 24x- 20y + 112 = 0
Locus of point P : x² + y² - 12x- 10y + 56 = 0