The circle c has equation $x^2 + y^2 - 2x + 8y + k = 0$, where $k \in \mathbb{R}$ - Leaving Cert Mathematics - Question 3 - 2022

To find the slope of the tangent to the circle at the point $(9, -4)$, we use implicit differentiation:

1. Differentiate both sides of the circle equation: $(x - 5)^2 + (y - 2)^2 = 20$

2. Applying the chain rule gives: $2(x - 5) + 2(y - 2) \frac{dy}{dx} = 0$

3. Isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x - 5}{y - 2}$

4. Substitute the point $(9, -4)$ into this equation: $\frac{dy}{dx} = -\frac{9 - 5}{-4 - 2} = -\frac{4}{-6} = \frac{2}{3}$

Thus, the slope of the tangent line at the point $(9, -4)$ is $\frac{2}{3}$.

To find the equation of the circles, we start with its center $(r, -r)$:

1. Using the point $(1, -8)$, we can find the relationship: $(1 - r)^2 + (-8 + r)^2 = r^2$

2. Expanding the equation gives: $(1 - r)^2 = 1 - 2r + r^2$ $(-8 + r)^2 = 64 - 16r + r^2$

3. Summing these gives: $1 - 2r + r^2 + 64 - 16r + r^2 = r^2$ $65 - 18r + r^2 = r^2$ $65 - 18r = 0\n \Rightarrow r = \frac{65}{18}$

Therefore, the equations of the circles can be derived as: