Question

Coordinates of the centre of a circle whose radius is $2$ units and touches the pair of lines $x^2-y^2-2x+1=0$ are

• A. $(1,2\sqrt{2})$
• B. $(1+2\sqrt{2},0)$
• C. $(1,-2\sqrt{2})$
• D. $(1-2\sqrt{2},0)$

Answer 1

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

$(1-2\sqrt{2},0)$

$x^2-y^2-2x+1=0\Rightarrow (x-1{)}^2-y^2=0\Rightarrow (x-1+y\left)\right(x-1-y)=0\Rightarrow x+y=1;x-y=1$
Hence centre of the required circle will lie on the angle bisector of above lines i.e., $y=0;x=1$

Case $1:$ When centre lies on $x=1$
Let centre be $(1,a)$
Then its distance from $x+y=1$ will be radius i.e., $2$ units
$\therefore \frac{|1+a-1|}{\sqrt{2}}=2\Rightarrow a=\pm 2\sqrt{2}$
Hence centres will be $(1,\pm 2\sqrt{2})$

Case $2:$ When centre lies on $y=0$
Let centre be $(b,0)$
Similar with case $1$
$\therefore \frac{|b-1|}{\sqrt{2}}=2\Rightarrow b=1\pm 2\sqrt{2}$
Hence centres will be $(1\pm 2\sqrt{2},0)$