Question

If $L$ represents a normal drawn at the point $P\left(\frac{\pi}{4}\right)$ on the circle $x^{2}+y^{2}+6x-6y-14=0$, then the equation of the diameter of this circle which is perpendicular to $L$ is

• A. $x-y+6=0$
• B. $2x+y+3=0$
• C. $3x+2y+3=0$
• D. $x+y=0$

Hint:

A line perpendicular to a normal line with slope $1$ has a slope of $-1$. Ensure it passes through the circle's center to qualify as a diameter.

Answer 1

The Correct Option is D

Step 1: Find the centre of the circle.
For $x^2+y^2+6x-6y-14=0$, the centre is $(-3,3)$ (half of the $x$ and $y$ coefficients, with sign changed).
Step 2: Understand the normal $L$.
A normal to a circle always passes through the centre. The point $P$ at parameter $\frac{\pi}{4}$ sits on the circle, and the normal joins the centre to this point.
Step 3: Find the slope of the normal.
At angle $\frac{\pi}{4}$, the point is $\big(-3+R\cos\frac{\pi}{4},\,3+R\sin\frac{\pi}{4}\big)$. The slope of the line from the centre to this point is $\tan\frac{\pi}{4}=1$. So $L$ has slope $1$.
Step 4: Slope of the wanted diameter.
The diameter we want is perpendicular to $L$. Perpendicular slope is the negative reciprocal of $1$, which is $-1$.
Step 5: A diameter passes through the centre.
Every diameter goes through the centre $(-3,3)$. So write a line with slope $-1$ through $(-3,3)$: \[ y-3=-1(x+3)\ \Rightarrow\ x+y=0. \]
Step 6: Match with the options.
The line $x+y=0$ has slope $-1$ and passes through the centre, so it is the required diameter. (The option $x-y+6=0$ has slope $1$, which is parallel to $L$, not perpendicular, so it is wrong.) \[ \boxed{x+y=0} \]