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Q.
If the circles $x^{2}+y^{2}-10x+16y+89-r^{2}=0$ and $x^{2}+y^{2}+6x-14y+42=0$ have common points, then the number of possible integral values of $r$ is equal to
NTA AbhyasNTA Abhyas 2022
Solution:
Centre & radius of $x^{2}+y^{2}-10x+16y+89-r^{2}=0$ are $\left(5 , - 8\right)$ & $\left|r\right|$ respectively
Centre & radius of $x^{2}+y^{2}+6x-14y+42=0$ are $\left(- 3,7\right)$ & $4$ respectively
Distance between the centres of circles is $\sqrt{8^{2} + 1 5^{2}}=17$
$\Rightarrow 17-4\leq \left|r\right|\leq 17+4$
$\Rightarrow 13\leq \left|r\right|\leq 21$
$\Rightarrow r\in \left\{\pm \, 21 , \pm 20 , \pm 19 , \pm 18 , \pm 17 , \pm 16 , \pm 15 , \pm 14 , \pm 13\right\}$
Hence, the number of possible values of $r$ is equal to $18$