JEE Main2002MathematicsCircleActual
The centres of a set of circles, each of radius 3 , lie on the circle x^2+y^2=25 . The locus of any point in the set is
Options
- A4 x^2+y^2 64
- Bx^2+y^2 25
- Cx^2+y^2 25
- D3 x^2+y^2 9
Correct answer
A. 4 x^2+y^2 64
Step-by-step solution
Let (h, k) be the centre of any such circle. Equation of such circle is (x-h)^2+(y-k)^2=3^2 . Since ( h , k ) lies on x ^2+ y ^2=25 h ^2+ k ^2=25 . x^2+y^2-(2 x h+2 y k)+25=9 ; Locus of (h, k) is x^2+y^2=16 , which clearly satisfies (a).