Let S be the set of all points left(frac{a}{b | vedclass

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(A) The equation of the circle is $x^2 + y^2 = 1$. Since the point $\left(\frac{a}{b}, \frac{c}{d}\right)$ lies on the circle,we have $\frac{a^2}{b^2}

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(A) The equation of the circle is $x^2 + y^2 = 1$. Since the point $\left(\frac{a}{b}, \frac{c}{d}\right)$ lies on the circle,we have $\frac{a^2}{b^2} + \frac{c^2}{d^2} = 1$,which implies $a^2 d^2 + c^2 b^2 = b^2 d^2$.Given that $b$ and $d$ are even,let $b = 2k$ and $d = 2m$ for some integers $k, m$. Since $\operatorname{HCF}(a, b) = 1$,$a$ must be odd. Similarly,since $\operatorname{HCF}(c, d) = 1$,$c$ must be odd.Substituting these into the equation: $a^2 (2m)^2 + c^2 (2k)^2 = (2k)^2 (2m)^2$,which simplifies to $4 a^2 m^2 + 4 c^2 k^2 = 16 k^2 m^2$.Dividing by $4$,we get $a^2 m^2 + c^2 k^2 = 4 k^2 m^2$.Since $a$ and $c$ are odd,$a^2$ and $c^2$ are odd. Thus,$a^2 m^2 + c^2 k^2$ is the sum of two terms. If $m$ and $k$ are both even,the equation is divisible by $4$,but the parity of the terms leads to a contradiction. If $m$ or $k$ are odd,the left side will be odd or have a different parity than the right side (which is a multiple of $4$).Specifically,$a^2 m^2 + c^2 k^2 \equiv m^2 + k^2 \pmod{4}$ (since odd squares are $\equiv 1 \pmod{4}$). For the sum to be $0 \pmod{4}$,both $m$ and $k$ would have to be even,which contradicts the condition that $b$ and $d$ are in simplest form with $a, c$ odd. Thus,no such integers exist.Therefore,the set $S$ is empty.

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