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If the circles $x^2+y^2-16 x-20 y+164=r^2(\mathrm{r}>0)$ and $x^2+y^2-8 x-14 y+29=0$ intersect in two distinct points, then the maximum possible integral value of $r$ is
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Verified Answer
The correct answer is:
10
We have circles
$$
\mathrm{S}_1 \equiv \mathrm{x}^2+\mathrm{y}^2-16 \mathrm{x}-20 \mathrm{y}+164-\mathrm{r}^2=0
$$
and $\mathrm{S}_2 \equiv \mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-14 \mathrm{y}+29=0$
Here $\mathrm{C}_1 \equiv(8,10)$ and $\mathrm{C}_2 \equiv(4,7)$
Now $\mathrm{C}_1 \mathrm{C}_2=5$, where radius of $\mathrm{S}_1=\mathrm{r}$
and radius of $\mathrm{C}_2=6$
Since both the circles intersect each other at two distinct points
hence, $\left|\mathrm{r}_1-\mathrm{r}_2\right| < \mathrm{C}_1 \mathrm{C}_2 < \mathrm{r}_1+\mathrm{r}_2$
$$
|\mathrm{r}-6| < 5 < \mathrm{r}+6
$$
On solving we get $r \in(1,11)$
Hence maximum integral value of $r=10$
$$
\mathrm{S}_1 \equiv \mathrm{x}^2+\mathrm{y}^2-16 \mathrm{x}-20 \mathrm{y}+164-\mathrm{r}^2=0
$$
and $\mathrm{S}_2 \equiv \mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-14 \mathrm{y}+29=0$
Here $\mathrm{C}_1 \equiv(8,10)$ and $\mathrm{C}_2 \equiv(4,7)$
Now $\mathrm{C}_1 \mathrm{C}_2=5$, where radius of $\mathrm{S}_1=\mathrm{r}$
and radius of $\mathrm{C}_2=6$
Since both the circles intersect each other at two distinct points
hence, $\left|\mathrm{r}_1-\mathrm{r}_2\right| < \mathrm{C}_1 \mathrm{C}_2 < \mathrm{r}_1+\mathrm{r}_2$
$$
|\mathrm{r}-6| < 5 < \mathrm{r}+6
$$
On solving we get $r \in(1,11)$
Hence maximum integral value of $r=10$
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