The lines $L_1$ and $L_2$ are parallel - AQA - A-Level Maths Pure - Question 8 - 2022 - Paper 1

Set up the relevant equations: Since both lines $L_1$ and $L_2$ are tangent to circle $C$, we can write the equations based on the mid-point of the line segment. The mid-point between the tangents can be expressed as: $\left( \frac{(0+10)}{2}, \frac{(5+\frac{11}{3})}{2} \right).$

Substitute the tangent conditions: For the point $(a, -17)$ to be a distance $r$ from the lines, equate the distances: The distance from point $(a, -17)$ to line $L_1$: $d_1 = \frac{|5a + 3(-17) - 15|}{\sqrt{5^2 + 3^2}} = \frac{|5a - 51 - 15|}{\sqrt{34}} = \frac{|5a - 66|}{\sqrt{34}}$

The distance from point $(a, -17)$ to line $L_2$: $d_2 = \frac{|5a + 3(-17) - 83|}{\sqrt{5^2 + 3^2}} = \frac{|5a - 51 - 83|}{\sqrt{34}} = \frac{|5a - 134|}{\sqrt{34}}$

Equate the distances: Since both distances are equal: $|5a - 66| = |5a - 134|\Rightarrow 5a - 66 = 5a - 134\text{ or } 5a - 66 = - (5a - 134).$ The first equation leads to a contradiction, while the second simplifies to: $10a = 198\Rightarrow a = 20.$

Conclusion: The value of $a$ is 20.

Center of the circle: The center of circle $C$ is given as $(20, -17)$.

Radius determined by tangents: The radius $r$ of the circle can be determined as half the distance from the two tangents: $r = \frac{d_1 + d_2}{2}$ Here both distances can also be computed from the distance formulas previously derived.

Write the standard equation of the circle: The standard form of the equation of circle is: $(x - a)^2 + (y - b)^2 = r^2$ Substituting $a = 20$ and $b = -17$ gives: $(x - 20)^2 + (y + 17)^2 = r^2$

Final equation of C: Substitute $r$ with the computed value to finalize the equation.