Thank you for uploading the image! I see a mathematical exercise involving tangents to circles. Let’s go through each question step-by-step to solve them. I'll start with Question 1 and proceed sequentially. Let me know if you'd prefer otherwise or need help on a specific problem first.

Question 1

The circle equation is:
$x^2 + y^2 - (a+2)x - 5ay + 19 = 0$ It passes through $P(-1, 6)$. Find the equation of the tangent to the circle at $P$.

Steps to Solve:

1. Verify if $P(-1, 6)$ satisfies the circle equation to confirm $P$ is on the circle: Substitute $x = -1, y = 6$ into the circle equation: $(-1)^2 + 6^2 - (a+2)(-1) - 5a(6) + 19 = 0$ Simplify: $1 + 36 + a + 2 - 30a + 19 = 0 \quad \Rightarrow \quad 58 - 29a = 0 \quad \Rightarrow \quad a = 2.$

2. Rewrite the circle equation with $a = 2$: Substituting $a = 2$ gives: $x^2 + y^2 - 4x - 10y + 19 = 0$

3. Find the gradient of the radius at $P(-1, 6)$: The center of the circle is obtained from the equation $x^2 + y^2 - 4x - 10y + 19 = 0$. Rewriting in standard form: $(x - 2)^2 + (y - 5)^2 = 25$ The center is $(2, 5)$. The gradient of the radius is: $m_{\text{radius}} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 6}{2 - (-1)} = -\frac{1}{3}.$

4. Find the gradient of the tangent: The tangent is perpendicular to the radius, so: $m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = 3.$

5. Equation of the tangent: Use the point-slope form of a line equation with $m = 3$ and point $(-1, 6)$: $y - 6 = 3(x + 1) \quad \Rightarrow \quad y = 3x + 9.$ Rearrange to standard form: $3x - y + 9 = 0.$

The correct answer is D: $3x - y + 9 = 0$.

Would you like me to proceed with Question 2, or clarify anything so far?