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Write the equation in standard form for the circle passing through ( - 1,3) centered at the origin.

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so we know the center of the circle is at 0,0, what's the radius?

well, the radius is the distance from the center to any point on the circle, it just so happen that -1, 3 is on it, so then


\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ 0 &,& 0~) % (c,d) &&(~ -1 &,& 3~) \end{array}~~~ % distance value d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ r=√((-1-0)^2+(3-0)^2)\implies r=√((-1)^2+3^2)\implies r=√(10)\\\\ -------------------------------


\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{0}{ h},\stackrel{0}{ k})\qquad \qquad radius=\stackrel{√(10)}{ r} \\\\\\ (x-0)^2+(y-0)^2=(√(10))^2\implies x^2+y^2=10
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User Vlado
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