A circle touches the y-axis at the point $$(0, 4)$$ and passes through the point $$(2, 0)$$. Which of the following lines is not a tangent to this circle?

We start by noting that the circle touches the y-axis at the point $$\,(0,4)\,.$$ Because the y-axis is the vertical line $$x=0,$$ tangency implies two things. First, the perpendicular from the centre to the y-axis is horizontal, so the centre must have the same y-coordinate as the point of contact; that is, the centre is $$\,(a,4)\,.$$ Second, the distance from the centre to the y-axis equals the radius, so the radius satisfies $$r=\lvert a\rvert.$$

The circle also passes through the point $$\,(2,0).$$ Using the distance formula between two points, the squared distance from the centre $$\,(a,4)$$ to this point must equal the squared radius:

$$\bigl(2-a\bigr)^2+\bigl(0-4\bigr)^2 = r^2.$$

We have already observed that $$r^2=a^2,$$ so substituting:

$$\bigl(2-a\bigr)^2+16=a^2.$$

Expanding the square and collecting like terms:

$$4-4a+a^2+16=a^2.$$

The $$a^2$$ terms cancel immediately, leaving

$$20-4a=0.$$

Solving for $$a$$ gives

$$4a=20 \;\;\Longrightarrow\;\; a=5.$$

Hence the centre is $$\,(5,4)$$ and the radius is $$r=\lvert5\rvert=5.$$ The standard form of the circle is therefore

$$\bigl(x-5\bigr)^2+\bigl(y-4\bigr)^2=25.$$

To determine whether a given line is tangent, we use the distance-from-a-point-to-a-line formula. For a line $$Ax+By+C=0,$$ the perpendicular distance from a point $$\,(x_0,y_0)$$ to the line is

$$d=\dfrac{\lvert Ax_0+By_0+C\rvert}{\sqrt{A^2+B^2}}.$$

The line is tangent to the circle precisely when this distance equals the radius $$r=5.$$ We now test each option using the centre $$\,(5,4).$$

Option A: $$4x-3y+17=0$$

Numerator: $$4(5)-3(4)+17=20-12+17=25.$$
Denominator: $$\sqrt{4^2+(-3)^2}=\sqrt{16+9}=5.$$
Distance: $$d=\dfrac{25}{5}=5=r.$$ So the line is a tangent.

Option B: $$3x-4y-24=0$$

Numerator: $$3(5)-4(4)-24=15-16-24=-25.$$
Denominator: $$\sqrt{3^2+(-4)^2}=\sqrt{9+16}=5.$$
Distance: $$d=\dfrac{\lvert-25\rvert}{5}=5=r.$$ So the line is a tangent.

Option C: $$3x+4y-6=0$$

Numerator: $$3(5)+4(4)-6=15+16-6=25.$$
Denominator: $$\sqrt{3^2+4^2}=\sqrt{9+16}=5.$$
Distance: $$d=\dfrac{25}{5}=5=r.$$ So the line is a tangent.

Option D: $$4x+3y-8=0$$

Numerator: $$4(5)+3(4)-8=20+12-8=24.$$
Denominator: $$\sqrt{4^2+3^2}=\sqrt{16+9}=5.$$
Distance: $$d=\dfrac{24}{5}=4.8\neq5.$$ The distance is not equal to the radius, so this line is not a tangent.

Hence, the correct answer is Option 4.