JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
A circle \(C\) of radius 2 lies in the second quadrant and touches both the coordinate axes. Let \(r\) be the radius of a circle that has centre at the point \((2,5)\) and intersects the circle \(C\) at exactly two points. If the set of all possible values of r is the interval \((\alpha, \beta)\), then \(3 \beta-2 \alpha\) is equal to :
- A \(10\)
- B \(15\)
- C \(12\)
- D \(14\)
Answer & Solution
Correct Answer
(B) \(15\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & C_1 C_2=\sqrt{(2+2)^2+(5-2)^2} \\ & =\sqrt{16+9} \\ & =5 \\ & r+2>5 \\ & r>3 \\ & r < 5+2 \\ & r < 7 \\ & \therefore \alpha=3, \beta=7 \\ & 3 \beta-2 \alpha=3(7)-2(3)=21-6=15\end{aligned}\)
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