Verified Solution • 2025–26 Session moderate

QJEEM22Maf183: JEE Main 2022 Mathematics - Class 11 JEEM22Maf183 Solution: If one of the diameters of the circle x 2 + y 2 − 2 2 x − 6 2 y …

If one of the diameters of the circle x2+y2−22x−62y+14=0 is a chord of the circle (x−22)2+(y−22)2=r2, then the value of r2 is equal to ____________.

Step-by-Step Verified Solution

For x2+y2−22x−62y+14=0x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0

 Radius =(2)2+(32)2−14=6

⇒\Rightarrow Diameter =26=2 \sqrt{6}

If this diameter is chord to

(x−22)2+(y−22)2=r2 then ⇒r2=6+((2)2+(2)2)2⇒r2=6+4=10⇒r2=10

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