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

The equation of the circle is $x^2 + y^2 - 2x + 4y - 15 = 0$ Rearranging gives: $x^2 + y^2 - 2x + 2(2)y - 15 = 0$ The center of this circle is (-1, -2) and the radius is: r = rac{ ext{sqrt}(b^2 - 4ac)}{2} where $g = -1$, $f = 2$, and $c = -15$. Thus, the radius is: $r = ext{sqrt}(1^2 + 2^2 - 15) = ext{sqrt}(20) = 2 ext{sqrt}(5)$ For the line to be tangent, the perpendicular distance from the line to the center of the circle must equal the radius. The formula for the distance from a point to a line is: rac{|Ax + By + C|}{ ext{sqrt}(A^2 + B^2)} Here, for the line $mx + 2y - 7 = 0$, we have: $A = m, B = 2, C = -7$ The distance from the point (-1, -2) is: rac{|m(-1) + 2(-2) - 7|}{ ext{sqrt}(m^2 + 2^2)} = rac{| -m - 4 - 7|}{ ext{sqrt}(m^2 + 4)} = rac{| -m - 11|}{ ext{sqrt}(m^2 + 4)} Setting this equal to the radius: rac{| -m - 11|}{ ext{sqrt}(m^2 + 4)} = 2 ext{sqrt}(5) Squaring both sides leads to solving a quadratic equation for $m$, which gives solutions: $m = 1 ext{ and } m = -41$