If the circles x^2+y^2-4x+6y+13-a^2=0 and x^2 | vedclass

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(D) The given equations of the circles are $x^2+y^2-4x+6y+13-a^2=0$ and $x^2+y^2-10x-2y+17=0$. For the first circle,the center $C_1 = (2, -3)$ and the

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(D) The given equations of the circles are $x^2+y^2-4x+6y+13-a^2=0$ and $x^2+y^2-10x-2y+17=0$. For the first circle,the center $C_1 = (2, -3)$ and the radius $r_1 = \sqrt{2^2 + (-3)^2 - (13-a^2)} = \sqrt{4+9-13+a^2} = |a|$. For the second circle,the center $C_2 = (5, 1)$ and the radius $r_2 = \sqrt{5^2 + 1^2 - 17} = \sqrt{25+1-17} = \sqrt{9} = 3$. The distance between the centers $d = C_1C_2 = \sqrt{(5-2)^2 + (1-(-3))^2} = \sqrt{3^2 + 4^2} = 5$. For the circles to intersect at two distinct points,the condition is $|r_1 - r_2| Substituting the values,we get $||a| - 3| From $5  2$,which implies $a \in (-\infty, -2) \cup (2, \infty)$. From $||a| - 3| Combining both conditions,$a \in (-8, -2) \cup (2, 8)$.

If the circles $x^2+y^2-4x+6y+13-a^2=0$ and $x^2+y^2-10x-2y+17=0$ intersect in two distinct points,then '$a$' is

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