Help Trigonometry!

Question

What are the solutions of the system

{y=x+3

{(x+1)²+(y-4)²=4

a. (1,4) and (-1,2)

b. (0,3) and (1,3)

Answer 1

I will define

y=x+3---(1)

(x+1)²+(y-4)²=4----(2)

Use the substitution method

(1) substitute to (2), Then

(x+1)²+(x+3-4)²=4

(x+1)²+(x-1)²=4

(x²+2x+1)+(x²-2x+1)=4 (expansion of (x+1)² and (x-1)²)

x²+2x+1+x²-2x+1=4 (remove parentheses)

2x²+2=4

x²=1

x²-1=0

(x+1)(x-1)=0

by zero product rule implies

x+1=0 or x-1=1

that is, x=-1 or x=1

Then use equation (1) to evaluate y

when x=-1, then y=-1+3=2

when x=1, then y=1+3=4

Therefore, the solutions of the system is {(-1,2), (1,4)}, That is (a)