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"Find the point at which the line touches the circle" I really need help no one helped last time...

The Line's equation is y=3x+10
The circle's equation is x^2+y^2=10

A. (-3,-1)
B. (3,1)
C. (-3,1)
D. No Solution

Respuesta :

the solution for a system of equations will be when the graphs touch each other, or intersect, and that happens when both equations equate each other, so, let's do so,

[tex]\bf \begin{cases} \boxed{y}=3x+10\\ -------\\ x^2+y^2=10\\ y^2=10-x^2\\ y=\sqrt{10-x^2} \end{cases}\qquad \qquad \boxed{3x+10}=\sqrt{10-x^2} \\\\\\ \textit{now we square both sides}\qquad (3x+10)^2=(\sqrt{10-x^2})^2 \\\\\\ (3x+10)^2=10-x^2\implies 9x^2+60x+100=10-x^2 \\\\\\ 10x^2+60x-90=0\implies x^2+6x-9=0\implies (x+3)(x+3)=0 \\\\\\ x= \begin{cases} -3\\ -3 \end{cases}\qquad thus\qquad \boxed{x=-3}[/tex]

[tex]\bf -------------------------------\\\\ \textit{now, when x = -3, what is \underline{y}?}\qquad y=3x+10\implies y=3(-3)+10 \\\\\\ y=-9+10\implies \boxed{y=1}\qquad \qquad therefore\qquad (\stackrel{x}{-3}~,~\stackrel{y}{1})[/tex]

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