Question

The total number of circles that can be drawn touching all the three lines $x+y-1=0$, $x-y-1=0,y+1=0$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

Given equation of lines are , $x+y=1$

$x-y=1$

$y=-1$

Intersection point of firt two lines is $(1,0)$. The third line does not pass through that point.

Hence, these are three non-collinear lines with different slopes.

So they will form a triangle.

We can have 4 circles touching all three sides of a triangle.

Three of them are exterior circles, and one is in-circle.