Write down, and simplify, the general term in the binomial expansion of $\left(x^2 + \frac{3}{x}\right)^{15}$. Hence determine the coefficient of $x^3$ in the binomial expansion of $\left(x^2 + \frac{3}{x}\right)^{15}$. Explain why there is no term in $\frac{1}{x^5}$ in the expansion of $\left(x^2 + \frac{3}{x}\right)^{15}$. The highest point on a second circle, $C_2$, is $(3,6)$ and the equation of the tangent at the lowest point is $y = 4$. Find the radius and the coordinates of the centre of $C_2$. Find the equation of $C_2$. Showing your working clearly, determine whether the circles $C_1$ and $C_2$ meet each other.

1. Write down the unsimplified general term using the binomial theorem

The general term $T_{r+1}$ is given by:
$T_{r+1} = \binom{15}{r} (x^2)^{15-r} \left(\frac{3}{x}\right)^r$

2. Simplify the general term by combining the powers of $x$

$T_{r+1} = \binom{15}{r} x^{30-2r} \cdot 3^r x^{-r}$
$T_{r+1} = \binom{15}{r} 3^r x^{30-3r}$

1. Find the general term of the expansion (from part i)

The general term is $T_{r+1} = \binom{15}{r} (x^2)^{15-r} \left(\frac{3}{x}\right)^r$
$T_{r+1} = \binom{15}{r} x^{30-2r} 3^r x^{-r}$
$T_{r+1} = \binom{15}{r} 3^r x^{30-3r}$

2. Equate the power of $x$ to 3 to find $r$

For the term in $x^3$, we set the exponent of $x$ to 3:
$30 - 3r = 3$
$3r = 27$
$r = 9$

3. Calculate the coefficient for $r = 9$

Substitute $r = 9$ into the coefficient part of the general term:
$\text{Coefficient} = \binom{15}{9} 3^9$
$= 5005 \times 19683$
$= 98513415$

1. Recall the general term from part (i) and equate its power of $x$ to $-5$

From part (i), the general term is $T_{r+1} = \binom{15}{r} (x^2)^{15-r} \left(\frac{3}{x}\right)^r = \binom{15}{r} 3^r x^{30-3r}$.
For the term in $\frac{1}{x^5} = x^{-5}$, we set the power of $x$ to $-5$:
$30 - 3r = -5$

2. Solve for $r$ and explain why the term does not exist

$3r = 35$
$r = \frac{35}{3} = 11\frac{2}{3}$
Since $r$ must be an integer ($0 \leqslant r \leqslant 15$), there is no integer value of $r$ that satisfies the equation. Therefore, there is no term in $\frac{1}{x^5}$.

1. Determine the $y$-coordinate of the lowest point and the diameter of the circle

The tangent at the lowest point is $y = 4$, so the lowest point has a $y$-coordinate of $4$.
The highest point is $(3, 6)$, so its $y$-coordinate is $6$.
The diameter is the vertical distance between the highest and lowest points:
Diameter $= 6 - 4 = 2$

2. Calculate the radius and the coordinates of the centre

Radius $r = \frac{\text{Diameter}}{2} = \frac{2}{2} = 1$.
The centre lies exactly halfway between the highest and lowest points. Since the highest point is $(3, 6)$ and the lowest point must share the same $x$-coordinate (as they lie on the vertical diameter), the lowest point is $(3, 4)$.
Centre $= \left(3, \frac{6+4}{2}\right) = (3, 5)$.

1. Use the centre and radius from the previous part to write the equation of the circle

From part (ii), the centre of $C_2$ is $(3, 5)$ and the radius is $1$.
The equation of the circle is:
$(x - 3)^2 + (y - 5)^2 = 1^2$
$(x - 3)^2 + (y - 5)^2 = 1$