点 $A(3, 8)$ 和 $B(6, 5)$ 在一个圆上。

1. Find the midpoint of $AB$

Midpoint $= \left(\frac{3+6}{2}, \frac{8+5}{2}\right) = (4.5, 6.5)$

2. Find the gradient of $AB$ and the perpendicular gradient

Gradient of $AB = \frac{5-8}{6-3} = \frac{-3}{3} = -1$

Gradient of perpendicular bisector $= -\frac{1}{-1} = 1$

3. Find the equation of the perpendicular bisector

$y - 6.5 = 1(x - 4.5)$
$y = x - 4.5 + 6.5$
$y = x + 2$

1. Find the equation of the perpendicular bisector of $AB$ (from part i)

Midpoint of $AB = \left(\frac{3+6}{2}, \frac{8+5}{2}\right) = (4.5, 6.5)$
Gradient of $AB = \frac{5-8}{6-3} = -1$
Gradient of perpendicular bisector $= 1$
Equation of perpendicular bisector: $y - 6.5 = 1(x - 4.5) \implies y = x + 2$

2. Solve simultaneously to find the center of the circle

The center lies on both the perpendicular bisector and the given line $y + x = 8$.
Substitute $y = x + 2$ into $y + x = 8$:
$(x + 2) + x = 8$
$2x = 6 \implies x = 3$
$y = 3 + 2 = 5$
Center is $(3, 5)$.

3. Calculate the radius of the circle

Radius is the distance from the center $(3, 5)$ to point $A(3, 8)$:
$r = \sqrt{(3-3)^2 + (8-5)^2}$
$r = \sqrt{0 + 3^2} = 3$

1. State the center and radius from part (ii) and substitute into the standard circle equation

From part (ii), the center is $(3, 5)$ and the radius is $3$.
The equation of the circle is:
$(x - 3)^2 + (y - 5)^2 = 3^2$
$(x - 3)^2 + (y - 5)^2 = 9$

1. Calculate the square of the distance from the center of the circle to point $C$.

From earlier parts, the center of the circle is $(3, 5)$ and the radius is $3$.
Let $d$ be the distance from the center to $C(6, 3)$.
$d^2 = (6 - 3)^2 + (3 - 5)^2$
$d^2 = 3^2 + (-2)^2$
$d^2 = 9 + 4 = 13$

2. Compare the distance to the radius to determine the position of $C$.

The square of the radius is $r^2 = 3^2 = 9$.
Since $d^2 = 13 > 9$, $d > r$.
Therefore, point $C$ is outside the circle.

1. Determine the distance from the center to the axes and compare with the radius

From part (ii), the center of the circle is $(3, 5)$ and the radius is $3$.
The perpendicular distance from the center to the $y$-axis is the $x$-coordinate of the center, which is $3$.
The perpendicular distance from the center to the $x$-axis is the $y$-coordinate of the center, which is $5$.

2. Conclude which axis is tangent to the circle

Since the distance from the center to the $y$-axis is exactly equal to the radius ($3 = 3$), the $y$-axis is tangent to the circle.