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The equation of the hypotenuse of an isosceles right angled triangle is $$x + 3y = 3.$$

The right angle is at the vertex $C(−2, 0)$.

(a) Find the two other vertices of the triangle.

(b) Find the equation of the circumscribed circle of the triangle.

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Let $\Delta ABC$ is a given triangle.

Hence, the altitude to $AB$ is $$\frac{|1\cdot(-2)+3\cdot0-3|}{\sqrt{1^2+3^2}}=\sqrt{\frac{5}{2}},$$ which gives $AC=BC=\sqrt5$.

Now, for finding $A$ and $B$ we need to solve a system: $(x+2)^2+y^2=5$ and $x=3-3y$, which gives $y^2-3y+2=0$ and $(0,1)$ and $(-3,2)$.

Thus, the middle point of $AB$ it's $\left(-\frac{3}{2},\frac{3}{2}\right)$ and the equation of the circle is $$\left(x+\frac{3}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{5}{2}$$ Done!

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