Question

alguien que me ayude con estos dos problemas por favor

1. Determinar la ecuación de la circunferencia que pasa por los puntos A (2,0), B (2,3), C(1, 3).


2. Hallar el centro y el radio de la circunferencia: x2 + y2 + 4x + 6y + 9 = 0.

Answer 1

$(x - k) {}^{2} + (y - h) {}^{2} = {r}^{2} \\ (2 - k) {}^{2} + (0 - h) {}^{2} = r {}^{2} \\ (2 - k) {}^{2} + (3 - h) {}^{2} = r {}^{2} \\ (1 - k) {}^{2} + (3 - h) {}^{2} = r {}^{2}$
$(3 - h) {}^{2} - ( - h) {}^{2} = 0 \\ 9 - 6h + h {}^{2} - h {}^{2} = 0 \\ h = \frac{3}{2}$
$(2 - k) {}^{2} - (1 - k) {}^{2} = 0 \\ 4 - 4k + {k}^{2} - 1 + 2k - {k}^{2} = 0 \\ 3 - 2k = 0 \\ k = \frac{3}{2}$
$(2 - \frac{3}{2} ) {}^{2} + ( - \frac{3}{2} ) {}^{2} = {r}^{2} \\ \frac{1}{4} + \frac{9}{4} = {r}^{2} \\ \frac{5}{2} = {r}^{2}$
Ecuación:
$(x - \frac{3}{2} ) {}^{2} + (y - \frac{3}{2} ) {}^{2} = \frac{5}{2} \\ {x}^{2} - 3x + \frac{9}{4} + {y}^{2} - 3y + \frac{9}{4} = \frac{5}{2} \\ {x}^{2} + {y}^{2} - 3x - 3y = - 2$