Question

If the distance of the point P from A(6,0) is twice its distance from the point B(1,3), prove that the locus of the point P is a circle . Also find its center and its radius.

Answer 1

Hint: In this question, we start by calculating the distance of point P from Point A and then calculate the distance of point P from Point B and then equate these two distances as per the given condition in the question $PA = 2PB$.

Complete step-by-step answer:
Let the point P be (x, y)
Now, according to the question
The distance of the point P from the point A(6,0) is twice its distance from the point B(1,3)
So, first of all, let us calculate the distance of point P from A(6,0) which is denoted as $PA$
$PA = \sqrt {{{(x - 6)}^2} + {{(y - 0)}^2}} $
And the distance of point P from B(1,3) is denoted by $PB$
 $PB = \sqrt {{{(x - 1)}^2} + {{(y - 3)}^2}} $
Now, according to the condition given in the question
$PA = 2PB$
$ \Rightarrow \sqrt {{{(x - 6)}^2} + {y^2}} = 2\sqrt {{{(x - 1)}^2} + {{(y - 3)}^2}} $
Squaring both sides , we get
$ \Rightarrow {(x - 6)^2} + {y^2} = 4[{(x - 1)^2} + {(y - 3)^2}]$
$ \Rightarrow {x^2} - 12x + 36 + {y^2} = 4[{x^2} - 2x + 1 + {y^2} - 6y + 9]$
$ \Rightarrow 3{x^2} + 3{y^2} + 4x - 24y + 4 = 0$
$ \Rightarrow {x^2} + {y^2} + \dfrac{4}{3}x - 8y + \dfrac{4}{3} = 0$
The above equation is the equation of the circle
${\text{Hence it is proved that the locus of the point P is a circle}}{\text{.}}$
Comparing the above equation with the standard equation of the circle which is
${x^2} + {y^2} + 2gx + 2fy + c = 0$
$\therefore {\text{On comparing, we get}}$
\[
  g = \dfrac{2}{3}, \\
  f = - 4, \\
  c = \dfrac{4}{3} \\
\]
The general form of circle’s center and radius is given below
$
  \therefore {\text{center}}( - g, - f) \\
   \Rightarrow {\text{center}}( - \dfrac{2}{3},4){\text{ }} {\text{ and}} \\
   \Rightarrow {\text{ radius}} = \sqrt {{g^2} + {f^2} - c} = \sqrt {\dfrac{{136}}{9}} = \dfrac{{2\sqrt {34} }}{3} \\
$

 
Note: The above given method is the best and easiest method to solve these types of questions. A locus is a set of all points that usually form a curve or surface and in the given question, we ended up with the equation of circle according to the condition mentioned in the question.