A line drawn through the point $$P(4, 7)$$ cuts the circle $$x^2 + y^2 = 9$$ at the points $$A$$ and $$B$$. Then $$P_A \cdot P_B$$ is equal to.

The circle is $$x^2 + y^2 = 9$$, so its centre is $$O(0,0)$$ and its radius is $$r = 3$$ because $$r^2 = 9$$.

The given point is $$P(4,7)$$. First find the square of the distance $$OP$$:

$$OP^2 = (4-0)^2 + (7-0)^2 = 4^2 + 7^2 = 16 + 49 = 65.$$

Clearly $$OP^2 = 65 > r^2 = 9$$, so the point $$P$$ lies outside the circle. Any line through an external point cuts the circle at two points; call them $$A$$ and $$B$$. We are required to compute $$PA \cdot PB$$.

Choose an arbitrary line through $$P$$. Let its slope be $$m$$. Its equation in point-slope form is

$$y - 7 = m\,(x - 4).$$

To find its intersections with the circle, substitute $$y = 7 + m(x - 4)$$ into the circle’s equation.

$$x^2 + \bigl[\,7 + m(x - 4)\bigr]^2 = 9.$$

Simplify the bracket:

$$7 + m(x - 4) = 7 + mx - 4m = mx + (7 - 4m).$$

Now square it:

$$\bigl(mx + (7 - 4m)\bigr)^2 = m^2x^2 + 2m(7 - 4m)x + (7 - 4m)^2.$$

Put this back into the circle’s equation:

$$x^2 + m^2x^2 + 2m(7 - 4m)x + (7 - 4m)^2 = 9.$$

Combine the like terms in $$x^2$$:

$$(1 + m^2)x^2 + 2m(7 - 4m)x + (7 - 4m)^2 - 9 = 0.$$

This quadratic equation in $$x$$ has roots $$x_1$$ and $$x_2$$ that correspond to the $$x$$-coordinates of $$A$$ and $$B$$. By Vieta’s formula, the product of the roots is

$$x_1x_2 = \dfrac{(7 - 4m)^2 - 9}{1 + m^2}.$$

Next, parameterise the line in a convenient way. Let $$t$$ measure distance along the line in the direction of increasing $$x$$. Write

$$x = 4 + t, \qquad y = 7 + mt.$$

For $$t = 0$$ we are at $$P(4,7)$$. The points $$A$$ and $$B$$ correspond to $$t = t_1$$ and $$t = t_2$$ where $$t_1$$ and $$t_2$$ are the solutions of the same quadratic, after substituting $$x = 4 + t$$ into the circle. Working this way is often faster, so proceed:

Substitute $$x = 4 + t$$ and $$y = 7 + mt$$ into the circle:

$$ (4 + t)^2 + \bigl(7 + mt\bigr)^2 = 9.$$

Expand:

$$ 16 + 8t + t^2 + 49 + 14mt + m^2t^2 = 9.$$

Group like terms:

$$ (1 + m^2)t^2 \;+\; (8 + 14m)t \;+\; (16 + 49 - 9) = 0.$$

Simplify the constant term:

$$16 + 49 - 9 = 56.$$

So the quadratic in $$t$$ is

$$ (1 + m^2)t^2 + (8 + 14m)t + 56 = 0.$$

Its two roots are $$t_1$$ and $$t_2$$ for points $$A$$ and $$B$$. Again by Vieta, their product is

$$t_1 t_2 = \dfrac{56}{1 + m^2}.$$

The distance between any two points on this line measured by the parameter $$t$$ must be converted to ordinary Euclidean distance. A displacement of $$\Delta t$$ along the line changes $$x$$ by $$\Delta t$$ and $$y$$ by $$m\,\Delta t$$, giving an actual length

$$\sqrt{(\Delta t)^2 + (m\,\Delta t)^2} = \sqrt{1 + m^2}\,|\Delta t|.$$

Therefore

$$PA = \sqrt{1 + m^2}\,|t_1|, \qquad PB = \sqrt{1 + m^2}\,|t_2|.$$

The required product is

$$PA \cdot PB = \bigl(\sqrt{1 + m^2}\bigr)^2\,|t_1 t_2| = (1 + m^2)\,\dfrac{56}{1 + m^2} = 56.$$

The factor $$1 + m^2$$ cancels, showing that the result is independent of the slope of the line. Thus for every possible line through $$P$$ cutting the circle, the constant product is

$$PA \cdot PB = 56.$$

This agrees with the well-known Power of a Point theorem, which states directly

$$PA \cdot PB = OP^2 - r^2 = 65 - 9 = 56.$$

Hence, the correct answer is Option C.