JEE Main 2022 · 28 Jun · Shift 2 · Mathematics Q23

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MathematicsInteger+4 / -1Question 23 of 90

If one of the diameters of the circle $x^2+y^2-2\sqrt2x-6\sqrt2y+14=0$ is a chord of the circle $\left(x-2\sqrt2\right)^2+\left(y-2\sqrt2\right)^2=r^2$ , then the value of $r^2$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 10

First circle. Centre $(\sqrt2,3\sqrt2)$ , radius $^2=2+18-14=6$ .

Diameter as chord. $r^2=6+$ (distance between centres) $^2=6+4=10$ .

Answer: 10

Topics:Equation of a circle

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JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 23