9.
[tex]3x + 2y - 4 = 0\\2y = -3x + 4\\y = -\frac{3}{2}x + 2[/tex]
[tex]a = -\frac{3}{2} \cdot (-3) + 2 = \frac{9}{2} + 2 = 4\frac{1}{2} + 2 = 6\frac{1}{2}[/tex]
[tex]\frac{1}{2} = -\frac{3}{2} \cdot 2b + 2 = -3b + 2\\3b = 2 - \frac{1}{2} = 1\frac{1}{2}\\b = 1\frac{1}{2} : 3 = \frac{3}{2} \cdot \frac{1}{3} = \frac{1}{2}[/tex]
[tex]-3c = -\frac{3}{2} \cdot (c - 2) + 2 = -\frac{3c}{2} + 3 + 2 = -\frac{3c}{2} + 5\\-6c = - 3c + 10\\10 = -6c + 3c = -3c\\c = \frac{10}{-3} = -3\frac{1}{3}[/tex]
10.
Niech punkt [tex]S(p,\ q)[/tex] oznacza środek okręgu.
[tex]p = \frac{-8 - 5}{2} = -\frac{13}{2} = -6\frac{1}{2}\\q = \frac{-2 + 3}{2} = \frac{1}{2}\\S = (-6\frac{1}{2},\ \frac{1}{2})[/tex]
Niech [tex]r[/tex] oznacza promień okręgu.
[tex](2r)^2 = 4r^2 = \left( 3 - (- 2) \right) ^2 + \left( -5 - (-8) \right)^2 = 5^2 + 3^2 = 25 + 9 = 34\\r^2 = 34 : 4 = 8\frac{1}{2}\\[/tex]
Równanie okręgu: [tex](x + 6\frac{1}{2})^2 + (y - \frac{1}{2})^2 = 8\frac{1}{2}[/tex]
