圆在点 $(-3,3)$ 处的切线通过原点。

1. Find the gradient of the tangent to the circle at $(-3, 3)$.

The tangent passes through $(-3, 3)$ and the origin $(0, 0)$.
Gradient of tangent, $m_t = \frac{3 - 0}{-3 - 0} = -1$

2. Find the gradient of the normal to the circle at $(-3, 3)$.

Since the normal is perpendicular to the tangent, its gradient $m_n$ satisfies $m_t \times m_n = -1$.
$m_n = \frac{-1}{-1} = 1$

3. Find the equation of the normal.

Using the point-slope form with point $(-3, 3)$ and gradient $m_n = 1$:
$y - 3 = 1(x - (-3))$
$y - 3 = x + 3$
$y = x + 6$

1. Find the equation of the first normal from part (a)

The tangent at $(-3, 3)$ passes through $(0, 0)$.
Gradient of tangent $= \frac{3 - 0}{-3 - 0} = -1$.
Gradient of normal $= -\frac{1}{-1} = 1$.
Equation of normal 1: $y - 3 = 1(x - (-3)) \implies y = x + 6$.

2. Find the equation of the second normal

The second normal is parallel to $5y = -2x + 7 \implies y = -\frac{2}{5}x + \frac{7}{5}$.
Gradient of normal 2 $= -\frac{2}{5}$.
It passes through $(0, -1)$, so its $y$-intercept is $-1$.
Equation of normal 2: $y = -\frac{2}{5}x - 1$.

3. Find the coordinates of the centre of the circle

The centre of the circle is the intersection of the two normals.
$x + 6 = -\frac{2}{5}x - 1$
$\frac{7}{5}x = -7 \implies x = -5$
$y = -5 + 6 = 1$
The centre is $(-5, 1)$.

4. Calculate the radius squared

The circle passes through $(-3, 3)$.
$r^2 = (-3 - (-5))^2 + (3 - 1)^2$
$r^2 = 2^2 + 2^2 = 8$

5. State the equation of the circle

Equation of the circle: $(x - (-5))^2 + (y - 1)^2 = 8$
$(x + 5)^2 + (y - 1)^2 = 8$

1. Find the center and radius of the circle from earlier parts

From (a), the normal at $(-3,3)$ has gradient $1$ (since tangent gradient is $-1$). Equation: $y = x + 6$.
From (b), the second normal has gradient $-\frac{2}{5}$ and passes through $(0,-1)$. Equation: $y = -\frac{2}{5}x - 1$.
Intersection of normals gives the center $(x_c, y_c)$:
$x_c + 6 = -\frac{2}{5}x_c - 1 \implies \frac{7}{5}x_c = -7 \implies x_c = -5$.
$y_c = -5 + 6 = 1$.
Center is $(-5, 1)$.
Radius $r = \sqrt{(-5 - (-3))^2 + (1 - 3)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.

2. Determine the coordinates of the point farthest from the y-axis

The distance of a point $(x,y)$ from the y-axis is $|x|$.
Since the entire circle lies in the region $x < 0$ (center at $x=-5$, radius $2\sqrt{2} \approx 2.83$), the point farthest from the y-axis is the one with the most negative x-coordinate.
This point lies horizontally to the left of the center.
$x$-coordinate $= x_c - r = -5 - 2\sqrt{2}$.
$y$-coordinate $= y_c = 1$.
The coordinates are $(-5 - 2\sqrt{2}, 1)$.