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Radius of Circle Touching Two Ellipses | JEE

JEE Maths question with a full step-by-step solution.

Question
A circle touches the major axes of the two ellipses x22086+y22010=1\dfrac{x^2}{2086}+\dfrac{y^2}{2010}=1 and x22010+y22086=1\dfrac{x^2}{2010}+\dfrac{y^2}{2086}=1 and one of their common tangents, with radius rr. Then [r][r] (greatest integer function) equals:
A1616
B1717
C1818correct
D1919
Solution
Step 1: The first ellipse has its major axis along the xx-axis; the second along the yy-axis. By symmetry a common tangent has slope m=±1m=\pm1. Take m=−1m=-1. Step 2: A tangent of slope −1-1 to the first ellipse is y=−x±2086⋅1+2010=−x±4096=−x±64y=-x\pm\sqrt{2086\cdot1+2010}=-x\pm\sqrt{4096}=-x\pm64. Take x+y=64x+y=64. Step 3: The circle touches both coordinate axes (the two major axes), so its centre is (r,r)(r,r). Tangency to x+y−64=0x+y-64=0 gives
∣2r−64∣2=r  ⇒  64−2r=r2  ⇒  r=642+2=32(2−2).\frac{|2r-64|}{\sqrt2}=r \;\Rightarrow\; 64-2r=r\sqrt2 \;\Rightarrow\; r=\frac{64}{2+\sqrt2}=32(2-\sqrt2).
 Step 4: Numerically r=32(2−2)≈18.745r=32(2-\sqrt2)\approx18.745, so [r]=18[r]=18. Correct answer: (3)
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