2016/101) The circles $x^2+y^2-10x + 16 = 0$ and $x^2+y^2 = r^2$ intersect each other in distinct points

1) $r < 2$ 2) $r > 8$ 3) $2 < r < 8$ 4) $2 <=r <= 8$

Solution
The circle $x^2+y^2 = r^2$ has centre at (0,0) and radius r
the circle $x^2+y^2 - 10x + 16 = 0$
$=>(x^2-10x + 25) + y^2 = 9$
$=>(x-5)^2 + y^2 = 3^2$
this has center at (5,0) and radius 3.
the distance between points = 5
one has radius 3 and another shall intersect if radius be between 5-3 and 5+ 3 or $2 < r < 8$ and hence(3)