Skip to main content
Back to 2D GEOMETRY SYSTEM

JEE MATHEMATICS MCQ: 2D GEOMETRY SYSTEM

1 Easy
1 of 1
Application +4 | -1
A point moves in such a way that the sum of square of its distance from the points A(2,0)A(2,0) and B(−2,0)B(-2,0) is always equal to the square of the distance between AA and BB. The locus of the point is
View answer and explanation
Correct option: D
x2+y2−4=0x^{2}+y^{2}-4=0
Option D (x2+y2−4=0x^{2}+y^{2}-4=0) is correct because the sum of the squares of the distances from (x,y)(x,y) to (2,0)(2,0) and (−2,0)(-2,0) being equal to the square of the distance between (2,0)(2,0) and (−2,0)(-2,0) leads to the equation (x−2)2+y2+(x+2)2+y2=16(x-2)^2 + y^2 + (x+2)^2 + y^2 = 16, which simplifies to x2+y2=4x^{2}+y^{2}=4 or x2+y2−4=0x^{2}+y^{2}-4=0.

More Questions from 2D GEOMETRY SYSTEM

Practice All Chapter MCQs →
Values of α\alpha if (α,2α)(\alpha, 2 \alpha) lies inside the ΔABC\Delta A B C if A(0,2),B(2,0)A(0,2), B(2,0) and C(4,4)C(4,4) The locus of the moving point PP, such that 2PA=3PB2 P A=3 P B where AA is (0,0)(0,0) and BB is (4,−3)(4,-3), is The incentre of the triangle with vertices (1,3),(0,0)(1, \sqrt{3}),(0,0) and (2,0)(2,0) is The triangle with vertex (1,5),(−3,1) & (−2,1)(1,5),(-3,1) \,\&\,(-2,1) is The area of the quadrilateral with its vertices at the foci of the conics 9x2−16y2−18x+32y−239 x^2-16 y^2-18 x+32 y-23 =0=0 and 25x2+9y2−50x−18y+33=025 x^2+9 y^2-50 x-18 y+33=0, is Orthocentre of the triangle whose vertices are (0,0)(3,0)(0,0)(3,0) and (0,4)(0,4) is
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →