Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - P.1 - Graphs and Models - Exercises: 62


The intersection points are $(x,y) = (-4,3)$ and $(x,y)=(-5,0)$.

Work Step by Step

The points of intersection are found by solving the system consisting of equations of the curves. In this problem we have \begin{align} &x^2+y^2=25\\ -&3x+y=15. \end{align} First express $y$ in terms of $x$ from the second equation: $$y=15+3x.$$ Then we substitute what we expressed in the first equation $$x^2+(15+3x)^2=25.$$ Expanding the bracket we get $$x^2+225+90x+9x^2=25.$$ We can arrange all of the nonzero terms on the left and leave zero to the right to get the quadratic equation in $x$: $$10x^2 + 90x +200=0.$$ We can divide the whole equation by $10$ to simplify: $$x^2+9x+20=0.$$ Now the solutions are $$x_{1,2} = \frac{-9\pm\sqrt{9^2-4\times20\times1}}{2} = \frac{-9\pm\sqrt{81-80}}{2} = \frac{-9\pm 1}{2}.$$ This evaluating for "plus" and for "minus" we get $$x_1=\frac{-9+1}{2} = -4;$$ $$x_2=\frac{-9-1}{2}=-5.$$ Now to find the points we have to find the $y$ coordinates. Returning $x_1$ and $x_2$ in the expression that determines $y$ in terms of $x$ we have $$y_1 = 3x_1+15= -12+15 = 3;$$ $$y_2=3x_2+15 = -15+15 = 0.$$ We get the points of intersection by arranging the $x$ and $y$ coordinates into pairs: $$(x,y)\in\{(-4,3),(-5,0)\}$$
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