The sides of a square are x = 2, x = 3, y = 1 | vedclass

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(A) The vertices of the square are formed by the intersection of the lines $x=2, x=3, y=1, y=2$. These vertices are $A(2, 1), B(3, 1), C(3, 2),$ and $

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$x^{2} + y^{2} - 5x - 3y + 8 = 0$

Vedclass · Answer

(A) The vertices of the square are formed by the intersection of the lines $x=2, x=3, y=1, y=2$. These vertices are $A(2, 1), B(3, 1), C(3, 2),$ and $D(2, 2)$. The diagonals of the square are $AC$ and $BD$. If we take the diagonal $AC$ as the diameter,the endpoints are $(2, 1)$ and $(3, 2)$. The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. Substituting the values for $AC$: $(x - 2)(x - 3) + (y - 1)(y - 2) = 0$. Expanding this: $x^2 - 5x + 6 + y^2 - 3y + 2 = 0$. Thus,$x^2 + y^2 - 5x - 3y + 8 = 0$. Similarly,if we take the diagonal $BD$ as the diameter,the endpoints are $(2, 2)$ and $(3, 1)$. The equation is $(x - 2)(x - 3) + (y - 2)(y - 1) = 0$,which results in the same equation: $x^2 + y^2 - 5x - 3y + 8 = 0$.

The sides of a square are $x = 2, x = 3, y = 1$ and $y = 2$. Find the equation of the circle drawn with the diagonal of the square as its diameter.

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