For 12:
Since the line y = mx + c is a tangent, then the distance from the centre of the circle to the line is equal to the radius.
Using perpendicular distance:
[tex]d = r = \frac{| -m \cdot a + 1 \cdot b - c |}{\sqrt{(-m)^{2} + 1}}[/tex]
[tex]r = \frac{|b + am - c|}{\sqrt{m^{2} + 1}}[/tex]
Squaring both sides, we get:
[tex]r^{2} = \frac{(b - c + ma)^{2}}{m^{2} + 1}[/tex]
[tex](1 + m^{2})r^{2} = (b - c + ma)^{2}[/tex], as required.
