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

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(A) The given equation of the circle is $x^2+y^2-ax-by=0$. Completing the square,we get: $(x^2-ax+\frac{a^2}{4})+(y^2-by+\frac{b^2}{4}) = \frac{a^2}{4

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given equation of the circle is $x^2+y^2-ax-by=0$. Completing the square,we get: $(x^2-ax+\frac{a^2}{4})+(y^2-by+\frac{b^2}{4}) = \frac{a^2}{4}+\frac{b^2}{4}$ $(x-\frac{a}{2})^2+(y-\frac{b}{2})^2 = (\frac{\sqrt{a^2+b^2}}{2})^2$. This is in the form $(x-h)^2+(y-k)^2=r^2$,where the center $(h, k) = (\frac{a}{2}, \frac{b}{2})$ and the radius $r = \frac{\sqrt{a^2+b^2}}{2}$. The parametric equations are given by $x = h + r \cos 	heta$ and $y = k + r \sin 	heta$. Substituting the values,we get $x = \frac{a}{2} + \frac{\sqrt{a^2+b^2}}{2} \cos 	heta$ and $y = \frac{b}{2} + \frac{\sqrt{a^2+b^2}}{2} \sin 	heta$.

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

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