Question

(x-3)^2+(y-3)^2=4 ; y=x-2​

Answer 1

Respuesta:

(x-3)^2+(y-3)^2=4   ;   y=x-2

Si no me equivoco, reemplazas el valor de "y" que es (x - 2) y resolvemos, entonces:

​$(x - 3)^2 + (y - 3)^2 = 4$

$(x-3)^2 + (x-2 - 3)^2 = 4\\\\(x^2 - 2(x)(3) + 3^2) + (x - 5)^2 = 4\\\\(x^2 - 6x + 9) + (x^2 - 2(x)(5) + 5^2) = 4\\\\x^2 - 6x + 9 + x^2 - 10x + 25 = 4\\\\2x^2 - 16x + 34 = 4\\\\2x^2 - 16x + 34 - 4 = 0\\\\2x^2 - 16x + 30 = 0\\\\\frac{(2x - 10)(2x - 6)}{2}\\\\\frac{2(x - 5)(2x - 6)}{2}\\\\(x - 5)(2x - 6)\\\\x-5 = 0\\x = 5\\\\\\2x - 6 = 0\\2x = 6\\x = 6/2\\x = 3$

Explicación paso a paso: