JEE MATHEMATICS MCQ: 2D GEOMETRY SYSTEM

1 Easy

1 of 1

Application +4 | -1

The coordinates of the point

A

and

B

are

(a k, 0)

and

\left(\frac{a}{k}, 0\right),(k= \pm 1)

. If a point

P

moves so that

P A=k P B

, then the equation to the locus of

P

View answer and explanation

Correct option: D

x^{2}+y^{2}-a^{2}=0

Option D (

x^2 + y^2 - a^2 = 0

) is correct because the equation represents a circle with radius 'a' centered at the origin, derived from the condition

PA = kPB

More Questions from 2D GEOMETRY SYSTEM

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\alpha

(\alpha, 2 \alpha)

\Delta A B C

A(0,2), B(2,0)

C(4,4)

P

2 P A=3 P B

A

(0,0)

B

(4,-3)

(1, \sqrt{3}),(0,0)

(2,0)

(1,5),(-3,1) \,\&\,(-2,1)

9 x^2-16 y^2-18 x+32 y-23

=0

25 x^2+9 y^2-50 x-18 y+33=0

(0,0)(3,0)

(0,4)