TS EAMCET · Maths · Circle
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
- A \(-8 < a < -2\)
- B \(a>8\)
- C \(a < -8\)
- D none of these
Answer & Solution
Correct Answer
(A) \(-8 < a < -2\)
Step-by-step Solution
Detailed explanation
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 \(\left|r_1-r_2\right| 2\) 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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