The parametric equations of the curve x^2+y^2 | vedclass

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(C) Given the equation of the circle: $x^2+y^2+ax+by=0$. Completing the square for $x$ and $y$ terms: $(x^2+ax+\frac{a^2}{4}) + (y^2+by+\frac{b^2}{4})

Vedclass · Accepted Answer

$x=-\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos 	heta, y=-\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin 	heta$

Vedclass · Answer

(C) Given the equation of the circle: $x^2+y^2+ax+by=0$. Completing the square for $x$ and $y$ terms: $(x^2+ax+\frac{a^2}{4}) + (y^2+by+\frac{b^2}{4}) = \frac{a^2}{4} + \frac{b^2}{4}$. This simplifies to: $(x+\frac{a}{2})^2 + (y+\frac{b}{2})^2 = \frac{a^2+b^2}{4}$. Comparing this with the standard form of a circle $(x-h)^2 + (y-k)^2 = r^2$,we identify the center $(h, k) = (-\frac{a}{2}, -\frac{b}{2})$ and the radius $r = \sqrt{\frac{a^2+b^2}{4}}$. The parametric equations for a circle with center $(h, k)$ and radius $r$ are given by $x = h + r \cos 	heta$ and $y = k + r \sin 	heta$. Substituting the values,we get $x = -\frac{a}{2} + \sqrt{\frac{a^2+b^2}{4}} \cos 	heta$ and $y = -\frac{b}{2} + \sqrt{\frac{a^2+b^2}{4}} \sin 	heta$.

The parametric equations of the curve $x^2+y^2+ax+by=0$ are

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