The line $m: 2x + 3y + 1 = 0$ is parallel to the line $n: 2x + 3y - 51 = 0$ - Leaving Cert Mathematics - Question 5 - 2018

Calculate the slope of line $m$:
The equation $m: 2x + 3y + 1 = 0$ can be rewritten in slope-intercept form as: $3y = -2x - 1 \Rightarrow y = -\frac{2}{3}x - \frac{1}{3}.$
Therefore, the slope of line $m$ is $-\frac{2}{3}$.

Find the slope of line $n$:
Since line $n$ is parallel to line $m$, it will have the same slope: $\text{slope of } n = -\frac{2}{3}.$

Formulate the equation of line $AB$ which is perpendicular to line $n$:
The slope of line $AB$ is the negative reciprocal of the slope of $n$:
$\text{slope of } AB = \frac{3}{2}.$
Using point $A(-2, 1)$ to find the equation: $y - 1 = \frac{3}{2}(x + 2) \Rightarrow y - 1 = \frac{3}{2}x + 3 \Rightarrow 2y - 2 = 3x + 6 \Rightarrow 3x - 2y + 8 = 0.$

Solve for intersection of lines $AB$ and $n$:
Set up the equations:

• Line $AB$: $3x - 2y + 8 = 0$
• Line $n$: $2x + 3y - 51 = 0$

Solving both simultaneously: From line $AB$, express $y$ in terms of $x$: $y = \frac{3}{2}x + 4.$ Substitute into line $n$:

Conclusion:
Therefore, the coordinates of point $B$ are $(6, 9)$.

Calculate the radius of circle $s$:
Let $h$ be the radius of circle $s$ and note the given ratio of radii: $r_s : r_t = 1 : 3.$
Thus, if $r_t = 3h$, since the distance between points $A$ and $B$ (on the tangents) is $AB = 8$, we have: $\frac{3h + h}{2} = 4 \Rightarrow 4h = 8 \Rightarrow h = 2.$

Identify the center of circle $s$:
Using the coordinates of $A(-2, 1)$ and the equation of tangents, the center $C$ (h,k) can be derived through geometric formulas. Due to geometric placement on line $m$, the coordinates become: $C = (x, y) = (-2, 1 + k)$
where $k = \sqrt{\frac{4}{3}} = \frac{4}{3}$. Thus: $Center = (-2, -1)$.

Equation of circle $s$:
Using the center and radius in the standard form: $(x + 2)^2 + (y + 1)^2 = (2)^2 = 4.$

Final equation:
Therefore, the equation of circle $s$ is:
$(x + 2)^2 + (y + 1)^2 = 4.$