Question

If $(a,\alpha )$ lies inside the circle $x^2+y^2=9;x^2-4x-a^2=0$ has exactly one root in $(-1,0)$, then find the area of the region in which $(a,\alpha )$ lies.

Answer 1

If $(a,\alpha )$ lies inside the circle

$x^2+y^2=9$ therefore

$a^2+{\alpha }^2<9$.......(1)

given $x^2-4x-a^2=0$ has exactly one root in $(-1,0)$

$f(-1)f\left(0\right)<0$

$-(5-a^2)a^2<0$

$(a^2-5)a^2<0$ $a^2$ always greater than zero

$a^2<5$......(2)

therefore ${\alpha }^2<4$

$|a|<\sqrt{5}|\alpha |<2$

$-\sqrt{5}<a<\sqrt{5}-2<\alpha <2$

Area of the region in which $(a,\alpha )$ lies is

$=4\times 2\sqrt{5}$

$=8\sqrt{5}$