JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let a circle \(C_1\) be obtained on rolling the circle \(x^2+y^2-4 x-6 y+11=0\) upwards 4 units on the tangent \(T\) to it at the point \((3,2)\). Let \(C _2\) be the image of \(C _1\) in T. Let A and B be the centers of circles \(C _1\) and \(C _2\) respectively, and \(M\) and \(N\) be respectively the feet of perpendiculars drawn from \(A\) and \(B\) on the \(x\)-axis. Then the area of the trapezium \(AMNB\) is :
- A \(2(2+\sqrt{2})\)
- B \(4(1+\sqrt{2})\)
- C \(3+2 \sqrt{2}\)
- D \(2(1+\sqrt{2})\)
Answer & Solution
Correct Answer
(B) \(4(1+\sqrt{2})\)
Step-by-step Solution
Detailed explanation
\(C =(2,3), r =\sqrt{2}\) \(\text { Centre of } G = A =2+4 \frac{1}{\sqrt{2}}\) \(3+\frac{4}{\sqrt{2}}=(2+2 \sqrt{2}, 3+2 \sqrt{2})\) \(A (2+2 \sqrt{2}, 3+2 \sqrt{2})\) \(B (4+2 \sqrt{2}, 1+2 \sqrt{2})\) \(\frac{ x -(2+2 \sqrt{2})}{1}=\frac{ y -(3+2 \sqrt{2})}{-1}=2\)…
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