A circle cuts the rectangular hyperbola ${\text{xy = 1}}$ in points $\left( {{{\text{x}}_{\text{r}}}{\text{,}}{{\text{y}}_{\text{r}}}} \right)$,${\text{r = 1,2,3,4}}$then
A.${{\text{x}}_{\text{1}}}{{\text{x}}_{\text{2}}}{{\text{x}}_{\text{3}}}{{\text{x}}_{\text{4}}}{\text { = - 1}}$
B.${{\text{x}}_{\text{1}}}{{\text{x}}_{\text{2}}}{\text{ + }}{{\text{x}}_{\text{3}}}{{\text{x}}_{\text{4}}}{\text{ = 1}}$
C.${{\text{x}}_{\text{1}}}{{\text{x}}_{\text{2}}}{{\text{x}}_{\text{3}}}{{\text{x}}_{\text{4}}}{\text{ = 1}}$
D.${{\text{x}}_{\text{1}}}{\text{ + }}{{\text{x}}_{\text{2}}}{\text{ + }}{{\text{x}}_{\text{3}}}{\text{ + }}{{\text{x}}_{\text{4}}}{\text{ = 0}}$

Hint: We can substitute the equation of the hyperbola in the equation of the circle. As they are intersecting, the points are the roots of the equation. Then we can find the product and sum of the roots from the equation and compare it with the given options.

Complete step by step answer:

We have the equation of parabola as ${\text{xy = 1}}$.
\[ \Rightarrow {\text{y = }}\dfrac{{\text{1}}}{{\text{x}}}\] … (1)
The equation of the circle is not given and as we don’t have the radius and center, we can write the equation as
\[{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + 2gx + 2fy + c = 0}}\]… (2)
Substituting (1) in (2), we get,
\[{{\text{x}}^{\text{2}}}{\text{ + }}{\left( {\dfrac{{\text{1}}}{{\text{x}}}} \right)^{\text{2}}}{\text{ + 2gx + 2f}}\dfrac{{\text{1}}}{{\text{x}}}{\text{ + c = 0}}\]
Multiplying throughout with ${{\text{x}}^{\text{2}}}$, we get,
\[{{\text{x}}^{\text{4}}}{\text{ + 1 + 2g}}{{\text{x}}^{\text{3}}}{\text{ + 2fx + c}}{{\text{x}}^{\text{2}}}{\text{ = 0}}\]
On rearranging, we get,
\[{{\text{x}}^{\text{4}}}{\text{ + 2g}}{{\text{x}}^{\text{3}}}{\text{ + c}}{{\text{x}}^{\text{2}}}{\text{ + 2fx + 1 = 0}}\]… (4)
Now we have a polynomial equation on degree 4. Its solutions will give the x coordinates points of intersection of the parabola and the circle.
For a 4th degree equation of the form \[{\text{a}}{{\text{x}}^{\text{4}}}{\text{ + b}}{{\text{x}}^{\text{3}}}{\text{ + c}}{{\text{x}}^{\text{2}}}{\text{ + dx + z = 0}}\], sum of the roots is given by, $\dfrac{{{\text{ - b}}}}{{\text{a}}}$and product is given by $\dfrac{{\text{z}}}{{\text{a}}}$.
So, the sum of the roots of equation (4) is given by, ${{\text{x}}_{\text{1}}}{\text{ + }}{{\text{x}}_{\text{2}}}{\text{ + }}{{\text{x}}_{\text{3}}}{\text{ + }}{{\text{x}}_{\text{4}}}{\text{ = }}\dfrac{{{\text{ - 2g}}}}{{\text{1}}}$
And product of the root is given by ${{\text{x}}_{\text{1}}}{{\text{x}}_{\text{2}}}{{\text{x}}_{\text{3}}}{{\text{x}}_{\text{4}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{1}}}{\text{ = 1}}$
So, the correct equation the roots satisfy is ${{\text{x}}_{\text{1}}}{{\text{x}}_{\text{2}}}{{\text{x}}_{\text{3}}}{{\text{x}}_{\text{4}}}{\text{ = 1}}$.
Therefore, the correct answer is option C.

Note: For a general polynomial \[{\text{P}}\left( {\text{x}} \right){\text{ = a}}{{\text{x}}^{\text{n}}}{\text{ + b}}{{\text{x}}^{{\text{n - 1}}}}{\text{ + c}}{{\text{x}}^{{\text{n - 2}}}}{\text{ + }}...{\text{ + z }}\], sum of the roots is given by $\dfrac{{{\text{ - b}}}}{{\text{a}}}$. For odd degree polynomials, i.e. n is odd, the product of the roots is $\dfrac{{{\text{ - z}}}}{{\text{a}}}$and for even degree polynomials, i.e. n is even, the product of the root is $\dfrac{{\text{z}}}{{\text{a}}}$.
Standard equation of a circle is given by ${\left( {{\text{x - }}{{\text{x}}_{\text{0}}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{y - }}{{\text{y}}_{\text{0}}}} \right)^{\text{2}}}{\text{ = }}{{\text{r}}^{\text{2}}}$, where r is the radius and $\left( {{{\text{x}}_{\text{0}}}{\text{,}}{{\text{y}}_{\text{0}}}} \right)$is the center. As we are taking the radius and center arbitrary, we take the expanded form of this equation. In this question, the sum of the roots becomes 0, when g becomes zero. We cannot choose this as a correct option as it is applicable only at certain conditions.