点 $A(8, 1)$ 和 $B(1, 2)$ 在以 $C$ 为圆心的圆上。直线 $4y = 3x - 20$ 在点 $A$ 处与圆相切。

1. Find the gradient of the tangent line

The equation of the tangent is $4y = 3x - 20$.
Rearranging into the form $y = mx + c$:
$y = \frac{3}{4}x - 5$
The gradient of the tangent, $m_t$, is $\frac{3}{4}$.

2. Find the gradient of the normal and its equation

The normal is perpendicular to the tangent, so its gradient $m_n$ is given by:
$m_n = -\frac{1}{m_t} = -\frac{4}{3}$
The normal passes through $A(8, 1)$. Using the point-gradient form:
$y - 1 = -\frac{4}{3}(x - 8)$
$3(y - 1) = -4(x - 8)$
$3y - 3 = -4x + 32$
$4x + 3y = 35$

1. Find the equation of the normal to the circle at $A$ (result from part i)

The tangent is $4y = 3x - 20$, so its gradient is $\frac{3}{4}$.
The gradient of the normal is $-\frac{4}{3}$.
Equation of normal at $A(8, 1)$:
$y - 1 = -\frac{4}{3}(x - 8)$
$3y - 3 = -4x + 32$
$4x + 3y = 35$
The centre $C(x, y)$ lies on this normal.

2. Form an equation using the fact that the distances from the centre to $A$ and $B$ are equal

Since $A(8, 1)$ and $B(1, 2)$ lie on the circle, $CA = CB$.
$(x - 8)^2 + (y - 1)^2 = (x - 1)^2 + (y - 2)^2$
$x^2 - 16x + 64 + y^2 - 2y + 1 = x^2 - 2x + 1 + y^2 - 4y + 4$
$-16x - 2y + 65 = -2x - 4y + 5$
$14x - 2y = 60$
$7x - y = 30 \implies y = 7x - 30$

3. Solve the simultaneous equations to find the coordinates of the centre $C$

Substitute $y = 7x - 30$ into the normal equation $4x + 3y = 35$:
$4x + 3(7x - 30) = 35$
$4x + 21x - 90 = 35$
$25x = 125 \implies x = 5$
$y = 7(5) - 30 = 5$
The centre $C$ is $(5, 5)$.

4. Calculate the radius of the circle

Radius $r = CA = \sqrt{(5 - 8)^2 + (5 - 1)^2}$
$r = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

5. Write the equation of the circle

Using the centre-radius form $(x - a)^2 + (y - b)^2 = r^2$:
$(x - 5)^2 + (y - 5)^2 = 5^2$
$(x - 5)^2 + (y - 5)^2 = 25$

1. State the centre and radius of the circle from part (ii)

From part (ii), the equation of the circle is $(x - 5)^2 + (y - 5)^2 = 25$.
The centre of the circle is $(5, 5)$ and the radius is $\sqrt{25} = 5$.

2. Explain the relationship between the centre's coordinates, the radius, and the coordinate axes

The perpendicular distance from the centre $(5, 5)$ to the $y$-axis is the $x$-coordinate, which is $5$.
The perpendicular distance from the centre $(5, 5)$ to the $x$-axis is the $y$-coordinate, which is $5$.
Since the distance to both axes is equal to the radius ($5$), both the $x$-axis and $y$-axis are tangents to the circle.