Question

The range of values of $\alpha$ for which the line $2y=gx+\alpha$ is a normal to the circle $x^{2}+y^{2}+2gx+2gy-2=0$ for all values of g is
A) $\left[ 1,\infty \right)$
B) $\left[ -1,\infty \right)$
C) (0,1)
D) $\left( -\infty ,1\right]$

Answer 1

Hint: In this question it is given that we have to find the range of values of $\alpha$ for which the line $2y=gx+\alpha$ is a normal to the circle $x^{2}+y^{2}+2gx+2gy-2=0$ for all values of g. So in order to find the range of $\alpha$ we have to put any point on the equation of the normal line. Since we know that the normal line of a circle always passes through the centre of a circle so by putting it we will get our condition.

Complete step-by-step solution:
So before moving into solution we have to know that if any equation of circle is in the form of $x^{2}+y^{2}+2gx+2fy+c=0$.......(1)
Then the coordinate of the centre is (-g,-f).
Given the equation of circle, $x^{2}+y^{2}+2gx+2gy-2=0$, so if we compare the given equation with equation (1) we can write the centre of the circle is (-g,-g).
Now since the line $2y=gx+\alpha$ passing through the centre (-g,-g) , so it must satisfy the equation, i.e,
$2\left( -g\right) =g\times \left( -g\right) +\alpha$
$\Rightarrow -2g=-g^{2}+\alpha$
$\Rightarrow g^{2}-2g-\alpha =0$..............(2)
As we know that if any quadratic equation $ax^{2}+bx+c=0$ is given then the quadratic formula is, $x=\dfrac{-b\pm \sqrt{b^{2}-4ac} }{2a}$
So by comparing this equation with equation(2), we get a=1, b=-2, c=$-\alpha$.
So by this formula the solution of equation (2) can be written as,
$g=\dfrac{-\left( -2\right) \pm \sqrt{\left( -2\right)^{2} -4\times 1\times \left( -\alpha \right) } }{2\times 1}$
$\Rightarrow g=\dfrac{2\pm \sqrt{4+4\alpha } }{2}$
$\Rightarrow g=\dfrac{2\pm 2\sqrt{1+\alpha } }{2}$
$\Rightarrow g=\dfrac{2\left( 1\pm \sqrt{1+\alpha } \right) }{2}$
$\Rightarrow g=1\pm \sqrt{1+\alpha }$
Now, since it is given that $g\in \left( -\infty ,\infty \right)$
i.e,$-\infty < g < \infty$,
By putting the value of g , we get,
$-\infty < 1\pm \sqrt{1+\alpha } < \infty$
$-\infty < \pm \sqrt{1+\alpha } < \infty$ [subtracting 1 from each side]
$-0 \leq 1+\alpha < \infty$ [by squaring]
$-1 \leq \alpha < \infty$ [subtracting 1 from each side]
Which can be written as $\alpha \in \left[ -1,\infty \right)$
Which is our required solution.
Hence the correct option is option B.

Note: While solving this type of problem you have to keep in mind that every normal line of a circle passes through the centre of the circle and this normal line also considered as diameter line that is, this line will intersect the circle in exactly two opposite points.