If (2, 0), (0, 1), (4, 5) and (0, c) are concyclic and then find c.

1.	If (2, 0), (0, 1), (4, 5) and (0, c) are concyclic and then find c.
Answer» x² + y² + 2gx + 2fy + c₁ = 0 Satisfies (2, 0), (0, 1) (4, 5) we get 4 + 0 + 4g + c₁ = 0 –– (i) 0 + 1 + 2g. 0 + 2f + c₁ = 0 –– (ii) 16 + 25 + 8g + 10f + c₁ = 0 –– (iii) (ii) – (i) we get – 3 – 4g + 2f = 0 4g – 2f = – 3 –– (iv) (ii) – (iii) we get – 40 – 8g – 8f = 0 (or) g + f = – 5 –– (v) Solving(iv) and (v) we get g = -13/6, f = -17/6 Substituting g and f value in equation (i) we get 4 + 4 (-13/6) + c₁ = 0 c₁ = 14/3 Now equation x² + y² – (13x/3) - (17y/3) + (14/3) = 0 Now circle passes through (0, c) then c² - (17c/3) + (14/3) = 0 3c² – 17c + 14 = 0 ⇒ (3c – 14) (c – 1) = 0 (or) c = 1 or 14/3

Answer»

x² + y² + 2gx + 2fy + c₁ = 0

Satisfies (2, 0), (0, 1) (4, 5) we get

4 + 0 + 4g + c₁ = 0 –– (i)

0 + 1 + 2g. 0 + 2f + c₁ = 0 –– (ii)

16 + 25 + 8g + 10f + c₁ = 0 –– (iii)

(ii) – (i) we get

– 3 – 4g + 2f = 0

4g – 2f = – 3 –– (iv)

(ii) – (iii) we get

– 40 – 8g – 8f = 0

(or) g + f = – 5 –– (v)

Solving(iv) and (v) we get

g = -13/6, f = -17/6

Substituting g and f value in equation (i) we get

4 + 4 (-13/6) + c₁ = 0

c₁ = 14/3

Now equation x² + y² – (13x/3) - (17y/3) + (14/3) = 0

Now circle passes through (0, c) then

c² - (17c/3) + (14/3) = 0

3c² – 17c + 14 = 0

⇒ (3c – 14) (c – 1) = 0

(or)

c = 1 or 14/3

Discussion

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