Q) A circle centered at (2, 1) passes through the points A (5, 6) and B (- 3, k) Find the value of k. Hence find length of chord.

(Q 26 A – 30/5/2 – CBSE 2026 Question Paper)

Ans:

(i) Value of k:

Let (2, 1) be the center O.

If points A and B lie on circle’s boundary, then OA = OB, being radii of the circle

We know that according to the distance formula, distance between 2 points (x₁,y₁) and (x₂,y₂)

D =

∴ Distance OA between O (2, 1) and A (5, 6)

OA =

∴ √(9 + 25) = √34

Now Distance OB between O (2, 1) and B (- 3, k)

OB =

∴ √(25 + (k ² – 2 k + 1))

∴ √(k ² – 2 k + 26)

Since OA = OB

∴ √34 = √(k ² – 2 k + 26)

By squaring on both sides, we get:

∴ 34 = k ² – 2 k + 26

∴ k ² – 2 k – 8 = 0

By mid term splitting:

∴ k ² – 4 k + 2 k – 8 = 0

∴ k (k – 4) + 2 (k – 4) = 0

∴ (k – 4) (k + 2) = 0

∴ k = 4 and k = – 2

Length of cord AB:

a) For k = 4, coordinates of point B are (- 3, 4)

Let this point be B₁

Now, Chord AB can be calculated by distance formula: D =

Here, AB₁ is the distance between A (5, 6) and B₁ (-3, 4)

∴ AB₁ =

∴ AB₁ = √68 = 2√17 units

b) For k = – 2, coordinates of point B are (-3, -2).

Let this point be B₂

Now, Chord AB can be calculated by distance formula: D =

Here, AB₂ is the distance between A (5, 6) and B₂ (- 3, – 2)

∴ AB₂ =

∴ AB₂ = √128 = 8√2 units

Therefore, for k = 4, chord AB is 2√17 units and for k = -2, chord AB is 8√2 units.

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