直线 $y=9$、$y=-1$ 和 $3y=4x+9$ 是一个圆的切线。 圆心的 x 坐标为正数。

1. Explain that the centre's y-coordinate is the midpoint of the two horizontal tangents.

The lines $y=9$ and $y=-1$ are horizontal tangents to the circle. The centre of the circle must lie exactly halfway between these two parallel tangents.

$y$-coordinate of centre $= \frac{9 + (-1)}{2} = \frac{8}{2} = 4$

1. Determine the radius of the circle

The circle is tangent to $y=9$ and $y=-1$. The distance between these horizontal tangents is the diameter.
Diameter $= 9 - (-1) = 10$
Radius $r = 5$

2. Form the equation of the circle with unknown x-coordinate of the centre

Let the centre of the circle be $(h, 4)$.
The equation of the circle is $(x - h)^2 + (y - 4)^2 = 5^2 = 25$.

3. Substitute the equation of the third tangent into the circle equation

The third tangent is $3y = 4x + 9$, so $x = \frac{3y - 9}{4}$.
Substitute $x$ into the circle equation:
$\left(\frac{3y - 9}{4} - h\right)^2 + (y - 4)^2 = 25$
$\left(\frac{3y - 9 - 4h}{4}\right)^2 + (y - 4)^2 = 25$
$\frac{(3y - (9 + 4h))^2}{16} + y^2 - 8y + 16 = 25$
$9y^2 - 6(9 + 4h)y + (9 + 4h)^2 + 16y^2 - 128y + 256 = 400$
$25y^2 - (54 + 24h + 128)y + (81 + 72h + 16h^2 + 256 - 400) = 0$
$25y^2 - (182 + 24h)y + (16h^2 + 72h - 63) = 0$

4. Apply the discriminant condition for a tangent

Since the line is a tangent, the discriminant of this quadratic in $y$ must be zero.
$b^2 - 4ac = 0$
$[-(182 + 24h)]^2 - 4(25)(16h^2 + 72h - 63) = 0$
$33124 + 8736h + 576h^2 - 100(16h^2 + 72h - 63) = 0$
$33124 + 8736h + 576h^2 - 1600h^2 - 7200h + 6300 = 0$
$-1024h^2 + 1536h + 39424 = 0$
Divide by $-1024$:
$h^2 - 1.5h - 38.5 = 0$
$2h^2 - 3h - 77 = 0$

5. Solve for h and select the positive value

$(2h + 11)(h - 7) = 0$
$h = -5.5$ or $h = 7$
Since the x-coordinate of the centre is positive, $h = 7$.

1. State the centre and radius of the circle using information from previous parts

From parts (a) and (b), the centre of the circle is $(7, 4)$.
The horizontal tangents are $y=9$ and $y=-1$, so the diameter is $9 - (-1) = 10$.
The radius of the circle is $r = \frac{10}{2} = 5$.

2. Calculate the distance from the centre to the point $(3,0)$ and compare it with the radius

Let $d$ be the distance from the centre $(7, 4)$ to the point $(3, 0)$.
$d^2 = (7 - 3)^2 + (4 - 0)^2$
$d^2 = 4^2 + 4^2 = 32$
Since the square of the radius is $r^2 = 5^2 = 25$, and $32 > 25$, the distance $d$ is greater than the radius $r$.
Therefore, the point $(3,0)$ lies outside the circle.