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

To find the coordinates of point B on line n which is closest to point A, we first determine the slope of line n. The equation of line n is given by:

$2x + 3y - 51 = 0$

Rearranging gives the slope of line n as $-\frac{2}{3}$. Since line AB is perpendicular to line n, the slope of line AB is the negative reciprocal of the slope of line n, which is:

$\frac{3}{2}.$

Next, we calculate the equation of line AB using point A and the slope we just found:

Using point-slope form:
$y - y_1 = m(x - x_1)\rightarrow y - 1 = \frac{3}{2}(x + 2)$

This simplifies to:

$y = \frac{3}{2}x + 4.$

Now, we can find the intersection point B between lines m and n by substituting the linear equation of line AB back into the equation of line n. This will allow us to solve for both x and y to find B.

To find the equation of circle s, we utilize the given radius ratio of 1:3 between circles s and t. Let the center of circle s be denoted as $C_s(h, k)$.

The radius of circle s can be expressed as:

$r_s = k,$

and for circle t, which has a given radius of
$\frac{3}{1}(r_s) = 3k.$

From here, we set up the formula for the distance from the point B to the center of circle s:

1. Distance Formula: The distance from $B$ to $(h, k)$ must equal the radius $r_s$:

$\sqrt{(h + 2)^2 + (k - 1)^2} = k.$

2. Equation of Circle s: Using the standard form for a circle, the equation will be:

$(x - h)^2 + (y - k)^2 = k^2.$

3. Substituting values: We can substitute the values for k to yield:

$(x + 1)^2 + (y - 2.5)^2 = (\frac{k}{\text{ratio constant}})^2.$

Thus, we will have a finalized circle equation for s based on our specific values of h and k which will encapsulate the geometric constraints given in the problem.