The circles x^2+y^2-10x+16=0 and x^2+y^2=a^2 | vedclass

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Question

(B) For two circles to intersect at two distinct points,the distance between their centers $d$ must satisfy the condition $|r_1 - r_2| < d < r_1 + r_2

Vedclass · Accepted Answer

$2 < a < 8$

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(B) For two circles to intersect at two distinct points,the distance between their centers $d$ must satisfy the condition $|r_1 - r_2| For the first circle $x^2+y^2-10x+16=0$,the center $C_1$ is $(5, 0)$ and the radius $r_1 = \sqrt{5^2 + 0^2 - 16} = \sqrt{25-16} = 3$. For the second circle $x^2+y^2=a^2$,the center $C_2$ is $(0, 0)$ and the radius $r_2 = |a|$. The distance between centers $d = \sqrt{(5-0)^2 + (0-0)^2} = 5$. Applying the condition $|r_1 - r_2| $|3 - |a|| From $5  2$,which implies $a > 2$ or $a From $|3 - |a|| Combining these,we get $2  0$,so $2 < a < 8$.

The circles $x^2+y^2-10x+16=0$ and $x^2+y^2=a^2$ intersect at two distinct points if

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