The circles $c_1$ and $c_2$ touch externally as shown - Leaving Cert Mathematics - Question 4 - 2013

To find the point of contact, we can divide the line segment joining the centres $(-3, -2)$ and $(1, 1)$ in the ratio of their radii, $2:3$.

Using the section formula, the coordinates are calculated as follows:

rac{(2 imes 1 + 3 imes -3)}{2 + 3}, rac{(2 imes 1 + 3 imes -2)}{2 + 3} = \left(\frac{(2 - 9)}{5}, \frac{(2 - 6)}{5}\right) = \left(-\frac{7}{5}, -\frac{4}{5}\right)

Thus, the coordinates of the point of contact are $\left(-\frac{7}{5}, -\frac{4}{5}\right)$.

To find the equation of the tangent, we first find the slope of the line joining the centres:

The slope from $(-3, -2)$ to $(1, 1)$ is:

$m_{centres} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-2)}{1 - (-3)} = \frac{3}{4}$

The tangent will be perpendicular to this line, hence the slope of the tangent $m$ will be:

$m = -\frac{4}{3}$

Using the point of contact $\left(-\frac{7}{5}, -\frac{4}{5}\right)$ and the slope in point-slope form, we have:

$y - \left(-\frac{4}{5}\right) = -\frac{4}{3}\left(x - \left(-\frac{7}{5}\right)\right)$

Rearranging gives:

$3y + 4x + 28/5 + 12/5 = 0 \implies 4x + 3y + 8 = 0$

Thus, the equation of the tangent $t$ is:

$4x + 3y + 8 = 0$