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A circle is drawn on the chord of a circle $x^{2}+y^{2}=a^{2}$ as diameter. The chord lies on the line $x+y=a$. What is the equation of the circle?
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The correct answer is:
$x^{2}+y^{2}-a x-a y=0$
Given, equation of circle $\Rightarrow x^{2}+y^{2}=a^{2}$ Equation of chord $\Rightarrow x+y=a$ (1) $\Rightarrow x^{2}+(a-x)^{2}=a^{2}$
$\Rightarrow x^{2}+a^{2}+x^{2}-2 a x=a^{2}$
$\Rightarrow 2 x^{2}=2 a x$
$\Rightarrow \mathrm{x}=0, \mathrm{a}$
When, $x=0, y=a$ and when $x=a, y=0$. $\therefore$ Points of intersection are $(0, \mathrm{a})$ and $(\mathrm{a}, 0)$ $\therefore$ Equation of circle with chord as diameter is $(x-0)(x-a)+(y-a)(y-0)=0$
$\Rightarrow x(x-a)+y(y-a)=0$
$\Rightarrow x^{2}-a x+y^{2}-a y=0$
$\Rightarrow x^{2}+y^{2}-a x-a y=0$
$\Rightarrow x^{2}+a^{2}+x^{2}-2 a x=a^{2}$
$\Rightarrow 2 x^{2}=2 a x$
$\Rightarrow \mathrm{x}=0, \mathrm{a}$
When, $x=0, y=a$ and when $x=a, y=0$. $\therefore$ Points of intersection are $(0, \mathrm{a})$ and $(\mathrm{a}, 0)$ $\therefore$ Equation of circle with chord as diameter is $(x-0)(x-a)+(y-a)(y-0)=0$
$\Rightarrow x(x-a)+y(y-a)=0$
$\Rightarrow x^{2}-a x+y^{2}-a y=0$
$\Rightarrow x^{2}+y^{2}-a x-a y=0$
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