Question

$ax + by + c = 0$ is the polar of $\left( {1,1} \right)$ with respect to the circle ${x^2} + {y^2} - 2x + 2y + 1 = 0$ and HCF of $a,b,c$ is equal to $1$, then find ${a^2} + {b^2} + {c^2}$.
A) $0$
B) $3$
C) $5$
D) $15$

Answer 1

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.