Question

Find the greatest integer values of a for which the point $(2a,a+1)$ is an interior point of the larger segment of the circle $x^2+y^2-2x-2y-8=0$ made by the chord whose equation is $x-y+1=0$.

Answer 1

$\because Point(2a,a+1)isaninteriorpointofcircle,\therefore (2a{)}^2+(a+1{)}^2-2\left(2a\right)-2(a+1)-8<0.\Rightarrow 4a^2+a^2+1+2a-4a-2a-2-8<0.\Rightarrow 5a^2-4a-9<0.\Rightarrow 5a^2-9a+5a-9<0.\Rightarrow 5a^2-5a-9a-9<0.\Rightarrow 5a(a+1)-9(a+1)<0.\Rightarrow (a+1)(a-\frac{9}{5})<0.a\epsilon (-1,\frac{9}{5}).\to \left(1\right).$

Point will lie on same side of line towards which centre of circle lie.

$\Rightarrow Centre\to (1,1).\Rightarrow x-y+1\Rightarrow 1-1+1=2>0.\therefore 2a-(a+1)+1>0.\Rightarrow 2a-a-1+1>0.\Rightarrow a>0\to \left(2\right).Using\left(1\right)\&\left(2\right):\Rightarrow a\epsilon (0,\frac{9}{5})$