Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 661: 16

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Given: $x^2+y^2=20\\y=x-4$ Substitute $y$ into the second equation: $x^2+(x-4)^2=20\\x^2+x^2-8x+16=0\\2x^2-8x+16=0\\x^2-4x+4$ Solving this we get $x_1=2+\sqrt 6\\x_2=2-\sqrt 6$ Find y: $y_1=x_1-4=-2+\sqrt 6\\y_2=x_2-4=-2-\sqrt 6$