A circle passes through $$(-2, 4)$$ and touches the y-axis at $$(0, 2)$$. Which one of the following equations can represent a diameter of this circle?

We are told that the required circle passes through the point $$(-2,\,4)$$ and that it touches the y-axis at the point $$(0,\,2)$$. “Touching the y-axis’’ means the y-axis is a tangent to the circle at that point.

Whenever a line is tangent to a circle, the radius drawn to the point of contact is perpendicular to the tangent. Because the y-axis is the line $$x=0$$, the normal (perpendicular) to it is a horizontal line, that is, a line parallel to the x-axis. Therefore the centre of the circle must lie somewhere on the horizontal line that passes through the point of tangency $$(0,\,2)$$. Hence we may write the centre as $$C(h,\,2)$$, where $$h$$ is unknown for the moment.

If a circle touches the y-axis, the distance from its centre to the y-axis equals the radius. The distance from the point $$C(h,\,2)$$ to the y-axis $$x=0$$ is simply $$|h|$$. So the radius of the circle is $$r = |h|$$.

Now we impose the second given condition: the circle must pass through the point $$(-2,\,4)$$. The distance between this point and the centre must therefore equal the radius. We write this condition using the distance formula.

Distance formula: the distance between $$(x_1,\,y_1)$$ and $$(x_2,\,y_2)$$ is $$\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.$$

Applying the formula between $$(-2,\,4)$$ and $$C(h,\,2)$$ gives $$\sqrt{\,(-2-h)^2 + (4-2)^2\,} = |h|.$$

Because every term under the square‐root sign is squared, we can remove the absolute value by squaring both sides.

Squaring both sides, we obtain

$$(-2-h)^2 + (4-2)^2 \;=\; h^2.$$

Simplify each part:

$$(-2-h)^2 = (h+2)^2 = h^2 + 4h + 4,$$ $$4-2 = 2 \quad\Longrightarrow\quad (4-2)^2 = 2^2 = 4.$$

Substituting these into the equation, we have

$$\bigl(h^2 + 4h + 4\bigr) + 4 \;=\; h^2.$$

Combine like terms on the left:

$$h^2 + 4h + 8 \;=\; h^2.$$

Subtract $$h^2$$ from both sides to eliminate the $$h^2$$ terms:

$$4h + 8 = 0.$$

Now solve for $$h$$:

$$4h = -8 \;\;\Longrightarrow\;\; h = -2.$$

So the centre of the circle is $$C(-2,\,2)$$ and the radius is $$r = |-2| = 2.$$

Any diameter of the circle must pass through this centre. Hence the correct equation for a diameter must be the equation of a straight line that contains the point $$(-2,\,2).$$ We now test each option.

Option A: $$2x - 3y + 10 = 0$$ Substitute $$(x,\,y) = (-2,\,2)$$: $$2(-2) - 3(2) + 10 = -4 - 6 + 10 = 0.$$ Since the result is $$0$$, this line indeed passes through the centre.

Option B: $$3x + 4y - 3 = 0$$ Substitute $$(x,\,y) = (-2,\,2)$$: $$3(-2) + 4(2) - 3 = -6 + 8 - 3 = -1 \neq 0.$$ This line does not pass through the centre.

Option C: $$4x + 5y - 6 = 0$$ Substitute $$(x,\,y) = (-2,\,2)$$: $$4(-2) + 5(2) - 6 = -8 + 10 - 6 = -4 \neq 0.$$ This line also misses the centre.

Option D: $$5x + 2y + 4 = 0$$ Substitute $$(x,\,y) = (-2,\,2)$$: $$5(-2) + 2(2) + 4 = -10 + 4 + 4 = -2 \neq 0.$$ This one likewise fails to pass through the centre.

Only Option A produces zero on substitution, so only Option A represents a line through the centre, and therefore only Option A can be a diameter of the given circle.

Hence, the correct answer is Option A.