The diagram shows two circles $c_1$ and $c_2$ of equal radius - Leaving Cert Mathematics - Question Question 1 - 2012

To verify if the point $P(-3, 4)$ lies on $c_1$, we substitute its coordinates into the equation of the circle:

$(-3)^2 + (4)^2 = 9 + 16 = 25$

Since the left side equals the right side, point $P(-3, 4)$ is indeed on $c_1$.

The center of $c_2$ is at $(-6, 8)$. Radius can be identified because the circles touch each other at point $P(-3, 4)$, and both have equal radii. The distance between $(0, 0)$ and $(-6, 8)$ is:

$ext{distance} = ext{sqrt}((-6-0)^2 + (8-0)^2) = ext{sqrt}(36 + 64) = ext{sqrt}(100) = 10$

The radius is half of this distance: $r = 10 / 2 = 5$

Thus, the equation for $c_2$ is: $(x + 6)^2 + (y - 8)^2 = 25$

To find the equation of the common tangent at point $P(-3, 4)$, we first determine the slope of the line joining the centers $(0, 0)$ and $(-6, 8)$:

ext{slope} = rac{8 - 0}{-6 - 0} = -rac{4}{3}

The slope of the tangent, being perpendicular to this line, is: ext{slope of the tangent} = rac{3}{4}

Using point-slope form of a line: $y - y_1 = m(x - x_1)$ Substituting $P(-3, 4)$ and the slope: y - 4 = rac{3}{4}(x + 3)

Rearranging: y - 4 = rac{3}{4}x + rac{9}{4} y = rac{3}{4}x + rac{25}{4}

Multiply through by 4 for standard form: $4y - 3x - 25 = 0$