Question

Coordinates of the points of intersection of the lines given by $L_1,L_2$ and $L_3$ with $x^2+y^2=5$ are

• A. $(\pm 1,\mp 2)$
• B. $(\pm 1/\sqrt{2},\pm 3\sqrt{2})$
• C. $(\pm \sqrt{5/2},\mp \sqrt{5/2})$
• D. $(\pm \sqrt{5/2},\pm \sqrt{5/2})$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $(\pm 1,\mp 2)$
B $(\pm 1/\sqrt{2},\pm 3\sqrt{2})$
D $(\pm \sqrt{5/2},\pm \sqrt{5/2})$

$L_1:6x^2+xy-y^2=0\Rightarrow (3x-y)(2x+y)=0$
If $2x+y$ coinsides with $L_2:3x^2-axy+y^2=0$

then $3+2a+4=0\Rightarrow a=-7/2$,
which is not possiable as $a>0$
So if $y=3x$ coinsides with $L_2$

then $3-3a+9=0\Rightarrow a=4$
$L_2:3x^2-4xy+y^2=0\Rightarrow (3x-y)(x-y)=0$

and $L_3:(2x+y)(x-y)=0$
$p_1=\pm \frac{16+4}{\sqrt{5}}=\pm 4\sqrt{5},p_2=\pm \frac{8-4}{\sqrt{2}}=\pm 2\sqrt{2}$
$\Rightarrow |p_1^2-p_2^2|=80-8=72$
Vertices of the triangle formed by the lines $L_3$ and $x=1$ are $(0,0),(1,-2).(1,1)$,

so the coordinates of the centroid of the triangle are $(2/3,-1/3)$.
the circle $x^2+y^2=5$ meets the line $3x-y=0$ at $(\pm 1/\sqrt{2},\pm 3\sqrt{2})$

the line $2x+y=0$ at $(\pm 1,\pm 2)$ and the line $x-y=0$ at $(\pm \sqrt{5/2},\pm \sqrt{5/2})$