Question

Let the equation $x^2-y^2-2x+2y=0$ represent the pair of tangents to the circle $2x^2-4x+2y^2+1=0$. If $a,L$ represent the lengths of tangents and equation of common chord respectively, then

• A. $L:2y=1$
• B. $a=1$ units
• C. $a=\frac{1}{\sqrt{2}}$ units
• D. $L:y=1$

Answer 1

Answer: (A), (C)

$a=\frac{1}{\sqrt{2}}$ units

$x^2-y^2-2x+2y=0\Rightarrow x^2-2x+1-(y^2-2y+1)=0\Rightarrow (x-1{)}^2-(y-1{)}^2=0\Rightarrow x+y=2;x=y$
So the intersection point is $(1,1)$
Now circle is $2x^2-4x+2y^2+1=0$
$\Rightarrow x^2-2x+y^2+\frac{1}{2}=0$
So the length of tangents will be $a=\sqrt{S_1}$
$\Rightarrow a=\sqrt{1-2+1+\frac{1}{2}}=\frac{1}{\sqrt{2}}$ units.
Now, equation of common chord will be $T=0$
$\therefore L:x-(x+1)+y+\frac{1}{2}=0\Rightarrow L:y=\frac{1}{2}$