Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Quiz for Lessons 9.6-9.7 - Page 664: 14

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From what we are given, we obtain $x^2+y^2=10^2\\(x-20)^2+(y-15)^2=15^2$ Multiply the second equation by $-1$ and add it to the first equation: $x^2-40x+400+y^2-30y+225=225\\(x^2+y^2)-40x-30y+400=0\\100-40x-30y+400=0\\-40x=30y-400\\x=12.5-\frac{3}{4}y$ Find the values of $y$ $(12.5-\frac{3}{4}y)^2+y^2=100\\156.25-18.75y+\frac{9}{16}y^2+y^2-100=0\\1.5625y^2-18.75y+56.25=0\\y=6$ Thus, $x=12.5-\frac{3}{4}(6)=8$