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ABCD is a square of side 'a'. If AB and AD are taken as co-ordinate axes, prove that the equation of the circle circumscribing the square is x^2+y^2=a(x+y). |
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Answer» SOLUTION :`THEREFORE` CENTRE of the circle is at (a/2, a/2) and radius is `(asqrt2)/2=a/sqrt2` `therefore` Equation of the circle is `(x-a/2)^2+(y-a/2)^2=a^2/2` or, `x^2+a^2/4-ax+y^2+a^2/4-ay=a^2/2` or, `x^2+y^2-ax-ay=0` or, `x^2+y^2=a(x+y)` |
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