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Question

(A) The sides of the rectangle are $x=4, x=-2, y=5, y=-2$. The vertices of the rectangle are the intersection points of these lines: $(4, 5), (-2, 5),

Vedclass · Accepted Answer

$x^2+y^2-2x-3y-18=0$

Vedclass · Answer

(A) The sides of the rectangle are $x=4, x=-2, y=5, y=-2$. The vertices of the rectangle are the intersection points of these lines: $(4, 5), (-2, 5), (-2, -2),$ and $(4, -2)$. The diagonal of the rectangle connects opposite vertices,for example,$(4, 5)$ and $(-2, -2)$. The equation of a 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 points $(4, 5)$ and $(-2, -2)$: $(x-4)(x+2) + (y-5)(y+2) = 0$ $x^2 + 2x - 4x - 8 + y^2 + 2y - 5y - 10 = 0$ $x^2 + y^2 - 2x - 3y - 18 = 0$.

The equation of the circle whose diameter is the diagonal of the rectangle with sides $x=4, x=-2, y=5, y=-2$ is:

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