Write down the equation of the circle with centre (–3, 2) and radius 4 - Leaving Cert Mathematics - Question 4 - 2012

First, rearrange the circle's equation as: $x^2 + y^2 - 2x + 4y - 15 = 0.$ Writing this in standard form yields the center and radius as follows: Center: $(-1, -2)$ and Radius: r = rac{1}{ oot 2 osup{c}}.

For the line to be tangent to the circle, the perpendicular distance from the line to the center must equal the radius. The distance from the line $mx + 2y - 7 = 0$ to the point $(-1, -2)$ is given by: $d = \frac{|m(-1) + 2(-2) - 7|}{\sqrt{m^2 + 2^2}}.$ Setting this equal to the radius, we get: $(|m + 4 - 7| = \frac{1}{\sqrt{2}}).$ This leads us to two forms to solve for $m$:

1. $|m - 3| = \frac{1}{\sqrt{2}}$, which gives us two cases:

• Case 1: $m - 3 = \frac{1}{\sqrt{2}}$
• Case 2: $m - 3 = -\frac{1}{\sqrt{2}}$ After solving both cases, the values of $m$ are:
  $m = 3 + \frac{1}{\sqrt{2}} \text{ and } m = 3 - \frac{1}{\sqrt{2}}.$