The circle passing through (1, -2) and touching the axis of $$x$$ at (3, 0) also passes through the point

We have a circle which passes through the fixed point $$(1,-2)$$ and is tangent to the $$x$$-axis at the point $$(3,0)$$. Because a circle that touches a line has its centre on the perpendicular drawn from the point of tangency to that line, the centre of our circle must lie somewhere on the vertical line $$x=3$$ passing through $$(3,0)$$.

If we denote the centre by $$(h,k)$$ and the radius by $$r$$, then for tangency to the $$x$$-axis we must have

$$h = 3,\qquad |k| = r.$$

This is simply the statement that the horizontal coordinate is fixed at $$3$$ and the vertical distance from the centre to the $$x$$-axis equals the radius.

So we set the centre as $$(3,k)$$ and the radius as $$r = |k|$$. Next, because the circle passes through the point $$(1,-2)$$, the distance from this point to the centre must equal the radius. Using the distance formula, which is

$$\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2},$$

we write

$$\sqrt{(1-3)^2 + \bigl(-2-k\bigr)^2}\;=\;|k|.$$

Squaring both sides (to remove the square root and the absolute value) gives

$$ (1-3)^2 + \bigl(-2-k\bigr)^2 \;=\; k^2. $$

Calculating term by term, we have

$$ (1-3)^2 = (-2)^2 = 4, $$

and

$$ \bigl(-2-k\bigr)^2 = (k+2)^2 = k^2 + 4k + 4. $$

Substituting these back, the equation becomes

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

Simplifying the left-hand side,

$$ 4 + k^2 + 4k + 4 = k^2. $$

Combining like terms we notice that $$k^2$$ appears on both sides and cancels out:

$$ 8 + 4k = 0. $$

Now solve for $$k$$:

$$ 4k = -8, \qquad\Rightarrow\qquad k = -2. $$

Since $$r = |k|$$, we find $$r = 2$$. Thus the centre of the circle is $$(3,-2)$$ and the radius is $$2$$.

With centre $$(3,-2)$$ and radius $$2$$, the circle’s equation is

$$ (x-3)^2 + (y+2)^2 = 2^2 = 4. $$

Now we test each of the four given options to see which point satisfies this equation.

Option A: $$(5,-2)$$

Substituting, $$ (5-3)^2 + (-2+2)^2 = 2^2 + 0^2 = 4. $$ Since the left-hand side equals the right-hand side, $$(5,-2)$$ lies on the circle.

Option B: $$(-2,5)$$

Substituting, $$ (-2-3)^2 + (5+2)^2 = (-5)^2 + 7^2 = 25 + 49 = 74 \neq 4. $$

Option C: $$(-5,2)$$

Substituting, $$ (-5-3)^2 + (2+2)^2 = (-8)^2 + 4^2 = 64 + 16 = 80 \neq 4. $$

Option D: $$(2,-5)$$

Substituting, $$ (2-3)^2 + (-5+2)^2 = (-1)^2 + (-3)^2 = 1 + 9 = 10 \neq 4. $$

Only Option A satisfies the equation of the circle.

Hence, the correct answer is Option A.