Find points on the y-axis which are at a distance of 10 units from the point (8, 8).

We know that,

x-coordinate of any point on y-axis = 0.

Let point on y-axis which is at a distance of 10 units from (8, 8) be P(0, y).

By distance formula,

$\Rightarrow d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \Rightarrow 10 = \sqrt{(0 - 8)^2 + (y - 8)^2} \\[1em] \Rightarrow 10 = \sqrt{(-8)^2 + y^2 + 8^2 - 16y} \\[1em] \Rightarrow 10 = \sqrt{64 + y^2 + 64 - 16y} \\[1em]

On squaring both sides,

$\Rightarrow 10^2 = 64 + y^2 + 64 - 16y \\[1em] \Rightarrow 100 = y^2 + 128 - 16y \\[1em] \Rightarrow y^2 - 16y + 128 - 100 = 0 \\[1em] \Rightarrow y^2 - 16y + 28 = 0 \\[1em] \Rightarrow y^2 - 14y - 2y + 28 = 0 \\[1em] \Rightarrow y(y - 14) - 2(y - 14) = 0 \\[1em] \Rightarrow (y - 2)(y - 14) = 0 \\[1em] \Rightarrow y - 2 = 0 \text{ or } y - 14 = 0 \\[1em] \Rightarrow y = 2 \text{ or } y = 14.$

∴ P = (0, y) = (0, 2) or (0, 14).

Hence, points on the y-axis which are at a distance of 10 units from the point (8, 8) are (0, 2) or (0, 14).