JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let the set of all values of r, for which the circles \((x+1)^{2}+(y+4)^{2}=r^{2}\) and \(x^{2}+y^{2}-4x-2y-4=0\) intersect at two distinct points be the interval \((\alpha, \beta)\). Then \(\alpha\beta\) is equal to
- A 25
- B 20
- C 21
- D 24
Answer & Solution
Correct Answer
(A) 25
Step-by-step Solution
Detailed explanation
\((x-2)^2+(y-1)^2=3^2 \&(x+1)^2+(y+4)^2=r^2\) \(\left| r _1- r _2\right|< c _1 c _2< r _1+ r _2\) \(|r-3|<\sqrt{(2+1)^2+(1+4)^2}\sqrt{34}\) \(-\sqrt{34}\sqrt{34}-3\) i.e.…
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