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?

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Answer 1 · 2018-07-31T17:26:37

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

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

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

It is given that, #PO^2+PA^2=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#.