If a circle of constant radius 3k passes thro | vedclass

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(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since the circle passes through the origin $(0, 0)$,$A(a, 0)$,and $B(0, b)$,the equatio

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$x^2 + y^2 = (2k)^2$

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(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since the circle passes through the origin $(0, 0)$,$A(a, 0)$,and $B(0, b)$,the equation of the circle is $x^2 + y^2 - ax - by = 0$.The radius of this circle is given by $\sqrt{(\frac{a}{2})^2 + (\frac{b}{2})^2} = 3k$.Squaring both sides,we get $\frac{a^2}{4} + \frac{b^2}{4} = 9k^2$,which implies $a^2 + b^2 = 36k^2$.Let $(x, y)$ be the centroid of $	riangle OAB$. Then $x = \frac{0+a+0}{3} = \frac{a}{3}$ and $y = \frac{0+0+b}{3} = \frac{b}{3}$.Thus,$a = 3x$ and $b = 3y$.Substituting these into the equation $a^2 + b^2 = 36k^2$,we get $(3x)^2 + (3y)^2 = 36k^2$.$9x^2 + 9y^2 = 36k^2$.Dividing by $9$,we get $x^2 + y^2 = 4k^2 = (2k)^2$.

If a circle of constant radius $3k$ passes through the origin $O$ and meets the coordinate axes at $A$ and $B$,then the locus of the centroid of the triangle $OAB$ is

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