Answer

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**Hint:**If ${x^2} + {y^2} - 2x + 2y + 1 = 0$ is the equation and we need to write the equation of polar of $\left( {{x_1},{y_1}} \right)$ with respect to this ${x^2} + {y^2} - 2x + 2y + 1 = 0$ is given as

$x{x_1} + y{y_1} - 2\left( {\dfrac{{x + {x_1}}}{2}} \right) + 2\left( {\dfrac{{y + {y_1}}}{2}} \right) = 0$ .

**Complete step-by-step answer:**

So question is saying that $ax + by + c = 0$ is the polar of $\left( {1,1} \right)$ with respect to the circle ${x^2} + {y^2} - 2x + 2y + 1 = 0$. So we know in polar form, we replace ${x^2}$ by $x{x_1}$, ${y^2}$ by $y{y_1}$,

$x$ by $\dfrac{{x + {x_1}}}{2}$ and $y$ by $\dfrac{{y + {y_1}}}{2}$

So, polar of $\left( {1,1} \right)$ with respect to the circle ${x^2} + {y^2} - 2x + 2y + 1 = 0$ is

$x{x_1} + y{y_1} - 2\left( {\dfrac{{x + {x_1}}}{2}} \right) + 2\left( {\dfrac{{y + {y_1}}}{2}} \right) = 0$

And here, $\left( {{x_1},{y_1}} \right) = \left( {1,1} \right)$

So putting ${x_1} = 1,{y_1} = 1$ we will get the equation.

$

x + y - \left( {x + 1} \right) + \left( {y + 1} \right) + 1 = 0 \\

2y + 1 = 0 \\

$

Here the equation of polar of $\left( {1,1} \right)$ with respect to the circle is given by

$2y + 1 = 0$

But it is given that $ax + by + c = 0$ is the polar of $\left( {1,1} \right)$ with respect to the circle.

So upon comparing equations $2y + 1 = 0$ and $ax + by + c = 0$

We get

$\dfrac{a}{0} = \dfrac{b}{2} = \dfrac{c}{1}$

So,

$b = 2,c = 1,a = 0$

As it is given that HCF of $a,b,c$ is equal to $1$, so, we found the values of $a,b,c$ that are equal to $0,2,1$ respectively.

Now according to question, it is asked to find ${a^2} + {b^2} + {c^2}$

$

= {0^2} + {2^2} + {1^2} \\

= 0 + 4 + 1 \\

= 5 \\

$

**Hence option C is correct.**

**Note:**We should know how to convert equation in polar of $\left( {{x_1},{y_1}} \right)$ just make some changes as follows:

replace ${x^2}$ by $x{x_1}$, ${y^2}$ by $y{y_1}$,

$x$ by $\dfrac{{x + {x_1}}}{2}$ and $y$ by $\dfrac{{y + {y_1}}}{2}$ and constant remains same.

HCF is one means its highest common factor will always be one.

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