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If one of the diameters of the circle x^2+y^2-2√2x -6√2y+14=0 is a chord of the circle (x-2√2)^2 +(y-2√2)^2= ^2, then the value of ^2 is equal to

JEE Main 2022 — Mathematics Coordinate Geometry

2022integermedium

If one of the diameters of the circle x2+y2−22x{x}^{2}+{y}^{2}-2\sqrt{2}xx2+y2−22​x −62y+14=0-6\sqrt{2}y+14=0−62​y+14=0 is a chord of the circle (x−22)2{(x-2\sqrt{2})}^{2}(x−22​)2 +(y−22)2=r2+{(y-2\sqrt{2})}^{2}={r}^{2}+(y−22​)2=r2, then the value of r2{r}^{2}r2 is equal to

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Held on 28 Jun 2022 · Verified 6 Jul 2026.

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