Question

Number of points which the graphs of $|y|=x+1$ and $(x-1{)}^2+y^2=4$ intersect, is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Let's first draw each of the graph separately

$\bullet \left|y\right|=x+1$

$\bullet {(x-1)}^2+y^2=z^2$

(Circle entered at $(1,0)$ having radius of $2$ units)

Combining both graphs we have

So we can clearly see that the two graphs intersect at three points, namely $(-1,0)$, $(0,1)$ & $(0,-1)$

$\therefore$ Answer : Option C.