JEE Advanced 2017 · Paper 1, Mathematics question 8

The circle. $(x+1)^2+(y+2)^2=5+p$: centre $(-1,-2)$, radius $r=\sqrt{5+p}$. It is at distance 1 from the $y$-axis and 2 from the $x$-axis.

Exactly three common points with the axes happens in two ways.

Through the origin ($p=0$, $r=\sqrt5$). It meets the $x$-axis at $0$ and $-2$, and the $y$-axis at $0$ and $-4$. That is 3 distinct points. ✓

Tangent to one axis, cutting the other.

• Tangent to the $x$-axis: $r=2$, so $p=-1$. It cuts the $y$-axis twice, since $1<2$. That is 3 points. ✓
• Tangent to the $y$-axis: $r=1$, so $p=-4$. It misses the $x$-axis, since $2>1$. Only 1 point.

So there are 2 values of $p$.

Answer: 2