A point moves in a manner that the sum of the squares of its distance from thr origin and the point (2,-3) is always 19.show that the locus of moving point is a circle.find the equation to the circle?

1 Answer
Jul 31, 2018

x^2+y^2-2x+3y-3=0x2+y22x+3y3=0.

Explanation:

Let the variable point be P(x,y)P(x,y) and the given fixed points be

O(0,0) and A(2,-3)O(0,0)andA(2,3).

It is given that, PO^2+PA^2=19PO2+PA2=19.

Using the distance formula, we have,

:. {(x-0)^2+(y-0)^2}+{(x-2)^2+(y+3)^2}=19.

:. 2x^2+2y^2-4x+6y+4+9=19, or,

x^2+y^2-2x+3y-3=0.

Rewriting after completing squares,

(x-1)^2+(y+3/2)^2=3+1+9/4=25/4=(5/2)^2.

This exhibits that the locus of the point is a circle having

centre at (1,-3/2) and radius 5/2.

It may be interesting to note that the centre of the circle is the

mid-point of OA.