Question

If the line y = mx does not intersect the circle $(x+10{)}^2+(y+10{)}^2=(6\sqrt{5}{)}^2,$ then prove that $-2<m<-1/2.$

Answer 1

Since the line does not intersect the given circle, p > r where p is perpendicular from centre (-10, -10) on the line y = mx
$\therefore \frac{-10(m-1)}{\sqrt{m^2+1}}>6sqrt5$
or 5 (m - 1) ${}^2$ > 9 (m${}^2$ + 1)
or 0 > 4m${}^2$ + 10m + 4 or 2m${}^2$ + 5m + 2 < 0
or (2m + 1)(m + 2) < 0 // 2 < m < -1/2
2nd Method: Eliminate y and find a quadratic in x and apply the condition that its roots are imaginary i.e., $\Delta$ <0.