1its a circle of radius \sqrt{\frac{2}{3}}a with its centre at the origin.
Ques) A circle circumscribing an equilateral triangle with centroid (0,0) of a side a is drawn and a square is drawn touching its four sides to circle. Then find the eqn of circle circumscribing the square.
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5 Answers
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1as triangle is equilateral, circumcentre coincides with centroid...
so centre of the first circle is (0,0) and radius is a/√3 .....
side of the square is 2a/√3...(draw the fig)
the centre of the 2nd circle is (0,0) and its diameter is the length of the square's diagonal =2√2a/√3......
1ok explanation goes as follows,
the radius of the circle circumscribing the triangle will be \frac{a}{\sqrt{3}}
now the radius of the bigger circle circumscribing the square will be \sqrt{2r^{2}}
where r is the radius of this smaller circle... hence u get the required result. and because of the triangle being equilaleteral , the symmetry of the centre is always there and (0,0) will be the centre of the required circle also.
19to make matters still simpler, think of the cube roots of unity lying on a unit circle...there |z| =1....now make your calculations easier !
and the answers given by jee2010 are absolutely correct !
