Question

The number of integral values of $k$ for which the line, $3x+4y=k$ intersects the circle, $x^2+y^2-2x-4y+4=0$ at two distinct points is

Answer 1

Find the integral values of $k$:

A equation of line $3x+4y=k$ and a equation $x^2+y^2-2x-4y+4=0$ of a circle is given.

Rewrite the equation of the circle as follows:

$$\begin{gathered} x^2+y^2-2x-4y+4+1-1=0 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y-2\right)}^2=1 \end{gathered}$$

Therefore, the centre of the given circle is $\left(1,2\right)$ and the radius is $1$.

Rewrite the equation of line as follows:

$3x+4y-k=0$

So, if the line intersects the circle at two distinct points then distance of centre from the line $<$ radius

$\frac{\left|3\left(1\right)+4\left(2\right)-k\right|}{\sqrt{3^2+4^2}}<1$.

$$\begin{gathered} \frac{\left|3+8-k\right|}{\sqrt{3^2+4^2}}<1 \\ \Rightarrow \frac{\left|11-k\right|}{5}<1 \\ \Rightarrow \left|11-k\right|<5 \\ \Rightarrow -5<k-11<5 \\ \Rightarrow 6<k<16 \end{gathered}$$

Therefore, the total number of integral values of $k$ is $9$.