Question:
The number of integral values of $\mathrm{k}$ for which the line, $3 x+4 y=k$ intersects the circle, $x^{2}+y^{2}-2 x-4 y+4=0$ at two distinct points is
Solution:
Circle $x^{2}+y^{2}-2 x-4 y+4=0$
$\Rightarrow(x-1)^{2}+(y-2)^{2}=1$
Centre: $(1,2)$ radius $=1$
line $3 x+4 y-k=0$ intersects the circle at two distinct points.
$\Rightarrow$ distance of centre from the line $<$ radius
$\Rightarrow\left|\frac{3 \times 1+4 \times 2-\mathrm{k}}{\sqrt{3^{2}+4^{2}}}\right|<1$
$\Rightarrow|11-\mathrm{k}|<5$
$\Rightarrow 6<\mathrm{k}<16$
$\Rightarrow \mathrm{k} \in\{7,8,9, \ldots \ldots 15\}$ since $\mathrm{k} \in \mathrm{I}$
Number of $K$ is 9