Point A(3, 4) is the center of a circle. If one of its diameters has one end as (7, 8); the other end of this diameter is :

1. (1, 0)

2. (0, 1)

3. (-1, 0)

4. (0, -1)

Given,

A(3, 4) is center and let B(7, 8) be one end of the diameter.

Let other end of diameter be C(x, y).

By mid-point formula,

Mid-point = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

From figure,

A is the mid-point of BC.

$\therefore (3, 4) = \Big(\dfrac{x + 7}{2}, \dfrac{y + 8}{2}\Big) \\[1em] \Rightarrow \dfrac{x + 7}{2} = 3 \text{ and } \dfrac{y + 8}{2} = 4 \\[1em] \Rightarrow x + 7 = 6 \text{ and } y + 8 = 8 \\[1em] \Rightarrow x = 6 - 7 \text{ and } y = 8 - 8 \\[1em] \Rightarrow x = -1 \text{ and } y = 0.$

∴ C = (-1, 0).

Hence, Option 3 is the correct option.