UPSC NDA I 2016 Mathematics Solved Paper

Concept:Two circles intersect at two distinct points when the distance between their centres is less than the sum of their radii and greater than the absolute difference of their radii.Explanation:Let the first circle be $(x-1)^2+(y-3)^2=r^2$.Its centre is $\mathrm{C_1}(1,3)$ and radius $r_1 = r$.The second circle is $x^2+y^2-8x+2y+8=0$.Rewrite as $(x-4)^2+(y+1)^2 = 9$.Its centre is $\mathrm{C_2}(4,-1)$ and radius $r_2 = \sqrt{4^2+1^2-8}=3$.Distance between centres: $\mathrm{C_1C_2} = \sqrt{(4-1)^2+(-1-3)^2} = \sqrt{9+16}=5$.For two distinct intersection points:$|r_1 - r_2| < \mathrm{C_1C_2} < r_1 + r_2$.Here $|r-3| < 5 < r+3$.From $5 < r+3$ we get $r > 2$.From $|r-3| < 5$ we get $-5 < r-3 < 5$, i.e. $-2 < r < 8$.Combining with $r > 0$ (radius), the condition is $2 < r < 8$.Answer:$2 < r < 8$ — Option D is correct.