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 ___________.

We have the circle

$$x^{2}+y^{2}-2x-4y+4=0$$

First, we put this equation into centre-radius form. We add and subtract the appropriate constants:

$$x^{2}-2x+1+y^{2}-4y+4=0+1+4$$

$$\bigl(x^{2}-2x+1\bigr)+\bigl(y^{2}-4y+4\bigr)=1$$

$$(x-1)^{2}+(y-2)^{2}=1$$

Thus the centre of the circle is $$C(1,\,2)$$ and the radius is $$r=1$$.

The straight line is

$$3x+4y=k.$$

For the line to cut the circle at two distinct points, the perpendicular distance from the centre of the circle to the line must be strictly less than the radius. We now state the distance formula:

For a line $$ax+by+c=0$$ and a point $$(x_{0},y_{0})$$, the perpendicular distance is

$$\displaystyle d=\frac{|ax_{0}+by_{0}+c|}{\sqrt{a^{2}+b^{2}}}\,.$$

First we rewrite the given line in the required form:

$$3x+4y-k=0,$$

so $$a=3,\;b=4,\;c=-k.$$ Substituting the centre $$(1,2)$$ into the formula gives

$$d=\frac{|3(1)+4(2)-k|}{\sqrt{3^{2}+4^{2}}} =\frac{|3+8-k|}{\sqrt{9+16}} =\frac{|11-k|}{5}.$$

For two distinct intersection points we need

$$d<r \;\; \Longrightarrow \;\; \frac{|11-k|}{5}<1.$$

Multiplying both sides by $$5$$ we get

$$|11-k|<5.$$

Now we convert the modulus inequality into a double inequality:

$$-5<11-k<5.$$

Subtracting $$11$$ everywhere,

$$-16<-k<-6.$$

Multiplying through by $$-1$$ (and reversing the inequality signs) gives

$$6<k<16.$$

The integral values satisfying this are

$$k=7,\,8,\,9,\,10,\,11,\,12,\,13,\,14,\,15.$$

Counting them, we find $$9$$ such integers.

So, the answer is $$9$$.