Question

If the line $x+2by+7=0$ is a diameter of the circle ${{x}^{2}}+{{y}^{2}}-6x+2y=0$ then $b$ is equal to
(a) $3$
(b) $-5$
(c) $-1$
(d) $5$

Answer 1

Hint: Compare the given equation of the circle from the general equation of the circle and find the centre of the circle. After that put the value of the centre in diameter equation to find the value of b.

Complete step-by-step answer:
We are given the equation of a circle that is,
${{x}^{2}}+{{y}^{2}}-6x+2y=0.......(i)$
We know the general equation of the circle is,
${{x}^{2}}+{{y}^{2}}+2gx+2fy+C=0........(ii)$
Comparing the equation (i) and (ii), we get
$2g=-6$ and $2f=2$
Dividing throughout by $'2'$ , we get
$g=\dfrac{-6}{2}$and $f=\dfrac{2}{2}$
$g=-3$$f=1$
In the general form of the equation of a circle the centre is $(-g,-f)$.
So, the centre coordinate of the given equation of circle is $(3,-1)$.
It has been told that the line is the diameter of the circle which means it passes through the circle’s centre only.
So, the coordinate $(3,-1)$must satisfy the equation of the line $x+2by+7=0$.
Putting the values of $x=3\text{ and }y=-1$, we get
$\Rightarrow x+2by+7=0$
$\Rightarrow 3+2b\times (-1)+7=0$
$\Rightarrow 3-2b+7=0$
$\Rightarrow 10=2b$
$\Rightarrow b=\dfrac{10}{2}$
$\therefore b=5$
So, option (d) is the correct answer.
Answer is option (d).

Note: We have to understand that the points or the coordinates which are coming as the centre of the circle must satisfy the given equation of line. So, you can put the value of x and y to calculate the value of ‘b’.
Students forget that the line is a diameter, so they may get confused that diameter passes through the centre.