Q. The locus of centre of a circle $x^{2} + y^{2} - 2x - 2y + 1 = 0$, which rolls outside the circle $x^{2} + y^{2} - 6x + 8y = 0$, is
A) $x^{2} + y^{2} - 2x - 2y - 34 = 0$
B) $x^{2} + y^{2} - 6x - 8y + 11 = 0$
C) $x^{2} + y^{2} - 6x + 8y - 11 = 0$
D) None of the above
Explanation
(c)
Centre of required circle $= (3,\ - 4)$
Radius of required circle $= 5 + 1 = 6$
[image:11_image49.emf]
$\therefore$ Locus of circle is
$(x - 3)^{2} + (y + 4)^{2} = 36$
$\Rightarrow x^{2} - 6x + 9 + y^{2} + 16 + 8y = 36$
$\Rightarrow x^{2} + y^{2} - 6x + 8y - 11 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.