Circle Question 7
Question 7 - 2024 (30 Jan Shift 1)
If the circles $(x+1)^{2}+(y+2)^{2}=r^{2} \quad$ and $x^{2}+y^{2}-4 x-4 y+4=0$ intersect at exactly two distinct points, then
(1) $5<\mathrm{r}<9$
(2) $0<\mathrm{r}<7$
(3) $3<r<7$
(4) $\frac{1}{2}<\mathrm{r}<7$
Show Answer
Answer (3)
Solution
If two circles intersect at two distinct points
$\Rightarrow\left|\mathrm{r}{1}-\mathrm{r}{2}\right|<\mathrm{C}{1} \mathrm{C}{2}<\mathrm{r}{1}+\mathrm{r}{2}$
$|r-2|<\sqrt{9+16}<r+2$
$|r-2|<5$ and $r+2>5$
$-5<\mathrm{r}-2<5 \quad \mathrm{r}>3$
$-3<r<7$
From (1) and (2)
$3<\mathrm{r}<7$
