Question

Find the equation of locus of a point which is at a distance 5 from A (4;-3)

Answer 1

Answer:

Locus of the point P is $x^2$ + $y^2$ - 8$x^2$ + 6y = 0

Step-by-step explanation:$x^2_1 + y^2_1-8x_1 + 6y_1 = 0$

Distance = AP = 5

⇒ $\sqrt{(x_{1}-4)^2 + (y_{1}+3)^2 = 5 }$

∴ Squaring on both sides,

= ${(x_{1}-4)^2 + (y_{1}+3)^2 = 25 }$

= $(x_1^2 + 16-8x_1)+(y^2_1+9+6y_1) = 25$

= $x^2_1+y^2_1-8x_1+6y_1=0$

∴ Equation of the Locus of P is $x^2$ + $y^2$ - 8$x^2$ + 6y = 0