Let $\alpha$ be an integer multiple of 8 . If 5 is the set or an possible values of $\alpha$ such that the…
Let $\alpha$ be an integer multiple of 8 . If 5 is the set or an possible values of $\alpha$ such that the line $6 x+8 y+\alpha=0$ intersects the circle $x^2+y^2-4 x-6 y+9=0$ at two distinct points, then the number of elements in S is
$4$
$6$
$2$
$1$
Solution
$x^2+y^2-4 x-6 y+9=0$
Radius $=\sqrt{4+9-9}=2 ;$ Centre $=(2,3)$
The line $6 x+8 y+\alpha=0$ intersects the circle at two distinct points
$\therefore$ Length of Perpendicular from centre $ \lt $ radius
$\Rightarrow\left|\frac{12+24+\alpha}{10}\right| \lt \Rightarrow-56 \lt \alpha \lt -16$
Since, $\alpha=8 k$ where $k \in \boldsymbol{Z}$
$\Rightarrow-56 \lt 8 k \lt -16 \Rightarrow-7 \lt k \lt -2$
Possible radius of $k=-3,-4,-5,-6$
$\therefore$ Number of elements in $S=4$