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 $$x - y + 1 = 0$$.

Solution

## $$\because \quad Point\quad (2a,a+1)\quad is\quad an\quad interior\quad point\quad of\quad circle,\\ \therefore \quad (2a{ ) }^{ 2 }+(a+1{ ) }^{ 2 }-2(2a)-2(a+1)-8<\quad 0.\\ \Rightarrow 4a^{ 2 }+{ a }^{ 2 }+1+2a-4a-2a-2-8<\quad 0.\\ \Rightarrow 5a^{ 2 }-4a-9<\quad 0.\\ \Rightarrow 5a^{ 2 }-9a+5a-9<\quad 0.\\ \Rightarrow 5a^{ 2 }-5a-9a-9\quad <\quad 0.\\ \Rightarrow 5a(a+1)-9(a+1)<0.\\ \Rightarrow (a+1)\quad (a-\cfrac { 9 }{ 5 } )\quad <0.\\ a\ \varepsilon \ (-1,\cfrac { 9 }{ 5 } ).\quad \rightarrow (1).$$Point will lie on same side of line towards which centre of circle lie.$$\Rightarrow Centre\quad \rightarrow (1,1).\\ \Rightarrow x-y+1\\ \Rightarrow 1-1+1=2>0.\\ \therefore \quad 2a-(a+1)+1>0.\\ \Rightarrow 2a-a-1+1>0.\\ \Rightarrow a>0\quad \rightarrow (2).\\ Using\quad (1)\quad \& \quad (2):\\ \Rightarrow a\varepsilon (0,\cfrac { 9 }{ 5 } )\quad$$Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More