If the circles $(x+1)^2+(y+2)^2=r^2$ and $x^2+y^2-4 x-4 y+4=0$ intersect at exactly two distinct points, then
Solution
<p>If two circles intersect at two distinct points</p>
<p>$$\begin{aligned}
& \Rightarrow\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{C}_1 \mathrm{C}_2<\mathrm{r}_1+\mathrm{r}_2 \\
& |\mathrm{r}-2|<\sqrt{9+16}<\mathrm{r}+2 \\
& |\mathrm{r}-2|<5 \text { and } \mathrm{r}+2>5 \\
& -5<\mathrm{r}-2<5 \quad \mathrm{r}>3 ~\text{......... 2}
\end{aligned}$$</p>
<p>$-3<\mathrm{r}<7\quad$ .... (1)</p>
<p>From (1) and (2)</p>
<p>$3<\text { r }<7$</p>
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
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