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Question: If the distance of the point P from A(6,0) is twice its distance from the point B(1,3), prove that t...

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.

Explanation

Solution

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=2PBPA = 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 PAPA
PA=(x−6)2+(y−0)2PA = \sqrt {{{(x - 6)}^2} + {{(y - 0)}^2}}
And the distance of point P from B(1,3) is denoted by PBPB
PB=(x−1)2+(y−3)2PB = \sqrt {{{(x - 1)}^2} + {{(y - 3)}^2}}
Now, according to the condition given in the question
PA=2PBPA = 2PB
⇒(x−6)2+y2=2(x−1)2+(y−3)2\Rightarrow \sqrt {{{(x - 6)}^2} + {y^2}} = 2\sqrt {{{(x - 1)}^2} + {{(y - 3)}^2}}
Squaring both sides , we get
⇒(x−6)2+y2=4[(x−1)2+(y−3)2]\Rightarrow {(x - 6)^2} + {y^2} = 4[{(x - 1)^2} + {(y - 3)^2}]
⇒x2−12x+36+y2=4[x2−2x+1+y2−6y+9]\Rightarrow {x^2} - 12x + 36 + {y^2} = 4[{x^2} - 2x + 1 + {y^2} - 6y + 9]
⇒3x2+3y2+4x−24y+4=0\Rightarrow 3{x^2} + 3{y^2} + 4x - 24y + 4 = 0
⇒x2+y2+43x−8y+43=0\Rightarrow {x^2} + {y^2} + \dfrac{4}{3}x - 8y + \dfrac{4}{3} = 0
The above equation is the equation of the circle
Hence it is proved that the locus of the point P is a 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
x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0
∴On comparing, we get\therefore {\text{On comparing, we get}}

g=23, f=−4, c=43 g = \dfrac{2}{3}, \\\ f = - 4, \\\ c = \dfrac{4}{3} \\\

The general form of circle’s center and radius is given below
∴center(−g,−f) ⇒center(−23,4)  and ⇒ radius=g2+f2−c=1369=2343  \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.