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If the extremities of a diameter of a circle are $(0,0)$ and
$\left(\mathrm{a}^{3}, 1 / \mathrm{a}^{3}\right)$, then the circle passes through which one of the
following points?
Options:
$\left(\mathrm{a}^{3}, 1 / \mathrm{a}^{3}\right)$, then the circle passes through which one of the
following points?
Solution:
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Verified Answer
The correct answer is:
$(1 / a, a)$
Given the extremities of a diameter of a circle as
$(0,0)$ and $\left(\mathrm{a}^{3}, \frac{1}{\mathrm{a}}\right)$, equation of circle is
$(x-0)\left(x-a^{3}\right)+(y-0)\left(y-\frac{1}{a^{3}}\right)=0$
$\Rightarrow \quad x^{2}-x a^{3}+y^{2}-\frac{y}{a^{3}}=0$
$\Rightarrow \quad x^{2}+y^{2}-x a^{3}-\frac{y}{a^{3}}=0$
Putting $\mathrm{x}=\frac{1}{\mathrm{a}}$ and $\mathrm{y}=\mathrm{a}$, the equation is satisfied.
Thus, the circle passes through the point $\left(\frac{1}{\mathrm{a}}, \mathrm{a}\right)$.
$(0,0)$ and $\left(\mathrm{a}^{3}, \frac{1}{\mathrm{a}}\right)$, equation of circle is
$(x-0)\left(x-a^{3}\right)+(y-0)\left(y-\frac{1}{a^{3}}\right)=0$
$\Rightarrow \quad x^{2}-x a^{3}+y^{2}-\frac{y}{a^{3}}=0$
$\Rightarrow \quad x^{2}+y^{2}-x a^{3}-\frac{y}{a^{3}}=0$
Putting $\mathrm{x}=\frac{1}{\mathrm{a}}$ and $\mathrm{y}=\mathrm{a}$, the equation is satisfied.
Thus, the circle passes through the point $\left(\frac{1}{\mathrm{a}}, \mathrm{a}\right)$.
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