Day: March 6, 2022

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202203061208 Solution to 1973-CE-AMATH-I-XX

A circle passes through the points (-1,-1), (3,3), and (7,-1). Find the equation of this circle.

Solution.

Let the centre of this circle be located at point (a,b) and its radius be R long. The equation of this circle is thus in the form

(x-a)^2+(y-b)^2=R^2,

where a, b, and R are three unknowns to be obtained in the following three equations:

\begin{aligned} (-1-a)^2+(-1-b)^2 & = R^2 \\ (3-a)^2+(3-b)^2 & = R^2 \\ (7-a)^2+(-1-b)^2 & = R^2 \\ \end{aligned}

\begin{aligned} (a^2+2a+1) + (b^2+2b+1) & = R^2 \\ (a^2-6a+9) + (b^2-6b+9) & = R^2 \\ (a^2-14a+49) + (b^2+2b+1) & = R^2 \\ \end{aligned}

\begin{aligned} a & = 3 \\ b & = -1 \\ R & = 4 \\ \end{aligned}

\therefore The equation of this circle is

(x-3)^2+(y+1)^2=16.