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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

\dfrac{x^2}{2086}+\dfrac{y^2}{2010}=1

and

\dfrac{x^2}{2010}+\dfrac{y^2}{2086}=1

and one of their common tangents, with radius

r

. Then

[r]

(greatest integer function) equals:

16

17

18

correct

19

Solution

Step 1: The first ellipse has its major axis along the

x

-axis; the second along the

y

-axis. By symmetry a common tangent has slope

m=\pm1

. Take

m=-1

. Step 2: A tangent of slope

-1

to the first ellipse is

y=-x\pm\sqrt{2086\cdot1+2010}=-x\pm\sqrt{4096}=-x\pm64

. Take

x+y=64

. Step 3: The circle touches both coordinate axes (the two major axes), so its centre is

(r,r)

. Tangency to

x+y-64=0

gives

\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-\sqrt2)\approx18.745

, so

[r]=18

. Correct answer: (3)

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