If one of the diameters of the circle $x^{2}+y^{2}-2 \sqrt{2} x -6 \sqrt{2} y+14=0$ is a chord of the circle $(x-2 \sqrt{2})^{2} +(y-2 \sqrt{2})^{2}=r^{2}$, then the value of $r^{2}$ is equal to

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Circle

If one of the diameters of the circle

x^{2} + y^{2} - 22 x - 62 y + 14 = 0

is a chord of the circle

(x - 22)^{2} + (y - 22)^{2} = r^{2}

, then the value of

r^{2}

is equal to

Answer: 10

Solution

The first circle has center

C (2, 32)

and radius

r_{1} = 2 + 18 - 14 = 6

. The second circle has center

O (22, 22)

. A diameter PQ of the first circle passes through C and is a chord of the second circle. The perpendicular from O to chord PQ passes through C (since PQ is a diameter of the first circle). In right triangle OCQ, where Q is an endpoint of the diameter on the first circle:

∣ O C ∣^{2} + ∣ C Q ∣^{2} = ∣ O Q ∣^{2}

. We have

∣ O C ∣^{2} = (22 - 2)^{2} + (22 - 32)^{2} = 2 + 2 = 4

and

∣ C Q ∣ = r_{1} = 6

. Thus

r^{2} = 4 + 6 = 10

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer

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