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Question

(A) Let the equation of the required circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.Since the circle passes through $(0,0)$,we have $(0-h)^{2}+(0-k)^{2}=r^{2}$

Vedclass · Accepted Answer

$x^{2}+y^{2}-ax-by=0$

Vedclass · Answer

(A) Let the equation of the required circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.Since the circle passes through $(0,0)$,we have $(0-h)^{2}+(0-k)^{2}=r^{2}$,which implies $h^{2}+k^{2}=r^{2}$.The equation of the circle becomes $(x-h)^{2}+(y-k)^{2}=h^{2}+k^{2}$.Since the circle makes intercepts $a$ and $b$ on the coordinate axes,it passes through $(a,0)$ and $(0,b)$.Substituting $(a,0)$ into the equation: $(a-h)^{2}+(0-k)^{2}=h^{2}+k^{2}$ $\Rightarrow a^{2}-2ah+h^{2}+k^{2}=h^{2}+k^{2}$ $\Rightarrow a^{2}-2ah=0$. Since $a 
eq 0$,we get $h=\frac{a}{2}$.Substituting $(0,b)$ into the equation: $(0-h)^{2}+(b-k)^{2}=h^{2}+k^{2}$ $\Rightarrow h^{2}+b^{2}-2bk+k^{2}=h^{2}+k^{2}$ $\Rightarrow b^{2}-2bk=0$. Since $b 
eq 0$,we get $k=\frac{b}{2}$.Substituting $h$ and $k$ back into the circle equation: $(x-\frac{a}{2})^{2}+(y-\frac{b}{2})^{2}=(\frac{a}{2})^{2}+(\frac{b}{2})^{2}$.Expanding this: $x^{2}-ax+\frac{a^{2}}{4}+y^{2}-by+\frac{b^{2}}{4}=\frac{a^{2}}{4}+\frac{b^{2}}{4}$.Simplifying,we get $x^{2}+y^{2}-ax-by=0$.

Find the equation of the circle passing through $(0,0)$ and making intercepts $a$ and $b$ on the coordinate axes.

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