Since (h,k) lies on circle x2+y2=4, we have h=2cosθ and k=2sinθ.
The transformed point is (2h+1,3k+2)=(4cosθ+1,6sinθ+2).
Rearranging: 16(x−1)2+36(y−2)2=1
This is an ellipse with a2=36, b2=16, so e2=1−a2b2=1−3616=95
Therefore e25=5/95=9
Let (h,k) lie on the circle C:x2+y2=4 and the point (2h+1,3k+2) lie on an ellipse with eccentricity e. Then the value of e25 is equal to ____.
Held on 24 Jan 2026 · Verified 6 Jul 2026.
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