Answer to Question #97315 in Analytic Geometry for Ojugbele Daniel

Question #97315
A square has verticles O(0,0), A(a,o), B(a,a) and C(o,a).
Find:
(a) the mid point of the diagonals OB and CA
(b) the length of a diagonal of the square and the radius of the circle in which OABC is inscribed
(c) the equation of the circle inscribing the square.
1
Expert's answer
2019-10-28T14:33:07-0400

Answer

(a) "D(\\frac{a}{2};\\frac{a}{2});"

(b) "OB=a\\sqrt{2};" "r=\\frac{a}{\\sqrt{2}};"

(c) "(x-\\frac{a}{2})^2+(y-\\frac{a}{2})^2=\\frac{a^2}{2}".

Explanation

(a) Because the vertices O(0,0), A(a,o), B(a,a) and C(o,a) are the vertices of the square, then o=0. The coordinates of the mid point the diagonals OB and CA are equal:

"x=\\frac{(0+a)}2 = \\frac{a}2", "y=\\frac{(0+a)}2 = \\frac{a}2." Thus, the mid point of the diagonals is "D(\\frac{a}2;\\frac{a}2)".

(b) The length of a diagonal of the square is equal to

"OB=CA=\\sqrt{(a-0)^2+(a-0)^2}=a\\sqrt{2}."

The radius of the circle in which OABC is inscribed will be

"r=\\frac{OB}2=\\frac{a}{\\sqrt{2}}."

(c) The equation of circle with center C(x0; y0) and radius r is written as

"(x-x_0)^2+(y-y_0)^2=r^2."

Because "D(\\frac{a}2;\\frac{a}2)" is the center of the circle in which OABC is inscribed, and the "r=\\frac{a}{\\sqrt{2}}" is its radius, then the equation of the circle inscribing the square is written as

"(x-\\frac{a}2)^2+(y-\\frac{a}2)^2=\\frac{a^2}2."


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