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If the circles $x^2+y^2-4 x+6 y+13-a^2=0$ and $x^2+y^2-10 x-2 y+17=0$ intersect in two distinct points, then ' $a$ ' is
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Verified Answer
The correct answer is:
$-8 < a < -2$
It is given that the circles
$x^2+y^2-4 x+6 y+13-a^2=0$
and $x^2+y^2-10 x-2 y+17=0$ intersect in two distinct points, the
$\left|r_1-r_2\right| < c_1 c_2 < r_1+r_2$
where, $c_1=(2,-3), c_2=(5,1), r_1=|a|$ and $r_2=3$
So, $\quad|| a|-3| < \sqrt{9+16} < |a|+3$
$\Rightarrow \quad|| a|-3| < 5 < |a|+3$
$\Rightarrow \quad|a|>2$ and $-5 < |a|-3 < 5$
So, $a \in(-\infty,-2) \cup(2, \infty)$ and $a \in(-8,8)$
Therefore, $a \in(-8,-2) \cup(2,8)$
$x^2+y^2-4 x+6 y+13-a^2=0$
and $x^2+y^2-10 x-2 y+17=0$ intersect in two distinct points, the
$\left|r_1-r_2\right| < c_1 c_2 < r_1+r_2$
where, $c_1=(2,-3), c_2=(5,1), r_1=|a|$ and $r_2=3$
So, $\quad|| a|-3| < \sqrt{9+16} < |a|+3$
$\Rightarrow \quad|| a|-3| < 5 < |a|+3$
$\Rightarrow \quad|a|>2$ and $-5 < |a|-3 < 5$
So, $a \in(-\infty,-2) \cup(2, \infty)$ and $a \in(-8,8)$
Therefore, $a \in(-8,-2) \cup(2,8)$
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