Question

$A(-2,0)$ and $B(2,0)$ are the two fixed points and $P$ is a point such that $PA-PB=2$. Let $S$ be the circle $x^2+y^2=r^2$, then match the following.

$\begin{array}{cc}ColumnI& ColumnII\\ \text{a. If}r=2,\text{then the number of points}P\text{satisfying}& \text{p. 2}\\ PA-PB=2\text{and lying on}{x}^2+y^2=r^2\text{is}& \\ \text{b. If}r=1,\text{then the number of points}P\text{satisfying}& \text{q. 4}\\ PA-PB=2\text{and lying on}{x}^2+y^2=r^2\text{is}& \\ \text{c. For}r=2\text{the number of common tangents is}& \text{r.}0\\ \text{d. For}r=\frac{1}{2}\text{the number of common tangents is}& \text{s.}1\end{array}$

• A. $a-q;b-p;c-r;d-p$
• B. $a-s;b-q;c-r;d-p$
• C. $a-p;b-s;c-r;d-p$
• D. $a-q;b-p;c-s;d-s$

Answer 1

The correct option is C $a-p;b-s;c-r;d-p$


The locus of point P satisfying $PA-PB=2$ is a branch of the hyperbola $x^2-\frac{y^2}{3}=1$.
For $r=2$, the circle and the branch of the hyperbola intersect at two points. For $r=1$, there is no point of intersection.
If $m$ is the slope of the common tangent, then
$m^2-3=r^2(1+m^2)$
or $m^2=\frac{r^2+3}{1-r^2}$
Hence, there are no common tangents for $r>1$ and two common tangents for $r<1$.