Question

The circle $x^2+y^2+x+y=0$ and $x^2+y^2+x-y=0$ intersect at an angle of

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

$x$ ${}^2$ +$y^2$ + $x$ + $y$ = $0$

$x$ ${}^2$ + $y^2$ + $x$ - $y$ = $0$

$2y=0$

$\Rightarrow$$y=0$

And $x$ ${}^2$ + 0 + $x$ +$0$ = $0$

$\Rightarrow$ $x$ ( $x$ + 1) = 0

$x$ = $0,-1$

$\therefore$ Points of intersection $\to (0,0)\&(1,0)$

$x$ ${}^2$ + $y^2$ + $x$ + y = 0

$\Rightarrow$ 2$x$ +$2y\frac{dy}{dx}$ + 1 + $\frac{dy}{dx}$ = 0

$\frac{dy}{dx}$ = $\frac{-2x-1}{2y+1}$

= $\frac{-(2x+1)}{2y+1)}$

$\Rightarrow m_1=-1=$ -1 (At (0,0)

At$(-1,0)$,$(${ m }_{ 1 }^{ 1 }$=$-(-1) $=1)$

$x$ ${}^2$ + $y^2$ + $x$ - $y$ =$0$

$\Rightarrow$ 2$x$ + 2y $\frac{dy}{dx}$ + 1 + $\frac{dy}{dx}$ = 0

$\frac{dy}{dx}=-\frac{(2x+1)}{2y-1}$

At(0,0)

$\frac{dy}{dx}=1=m_2$

$\because$ $m_1$ $m_2$ = -1

$\therefore$ Angle between them =$\frac{\pi }{2}$.