Question

[Key Concept: Points on $x^{2} + y^{2} = 9$ at distance $2$ from $P$ are the ends of a chord of length $2$ through $P$ — rotate $P$ by the corresponding central angle.]

Given circle: x² + y² = 9

Therefore, the centre of the circle is O(0, 0) and its radius is:

r = 3, Given point: P(1, 2√2)

We can verify that P lies on the circle:

1² + (2√2)² = 1 + 8 = 9

Let Q(x, y) be a point on the given circle such that PQ = 2.

Therefore, x² + y² = 9 and (x − 1)² + (y − 2√2)² = 4

Expanding the second equation:

x² + y² − 2x − 4√2y + 1 + 8 = 4

Using x² + y² = 9, 9 − 2x − 4√2y + 9 = 4

Therefore, 2x + 4√2y = 14 or x + 2√2y = 7

Now we check the given options.

Option (a)

Q = (−1, 2√2)

Distance from P:

PQ² = (−1 − 1)² + (2√2 − 2√2)²= (−2)²= 4

Therefore, PQ = 2 Also, (−1)² + (2√2)² = 1 + 8 = 9

Hence, (a) is correct.

Option (b)

Q = (2√2, 1)

PQ² = (2√2 − 1)² + (1 − 2√2)²= (2√2 − 1)² + (1 − 2√2)²= 2(2√2 − 1)²≠ 4

Hence, (b) is incorrect.

Option (c)

Q = (23/9, 10√2/9)

First, verify that Q lies on the circle:

x² + y²= (23/9)² + (10√2/9)²= 529/81 + 200/81= 729/81= 9

Now,

PQ² = (23/9 − 1)² + (10√2/9 − 2√2)²= (14/9)² + (−8√2/9)²= 196/81 + 128/81= 324/81= 4

Therefore,PQ = 2

Hence, (c) is correct.

Option (d)

Q = (3, 0)

PQ² = (3 − 1)² + (0 − 2√2)²= 4 + 8= 12

Therefore, PQ ≠ 2

Hence, (d) is incorrect.

Final Answer

Correct options: (a) and (c)