Circle — Maths STD 11 Science — Question

$x^2+y^2+2 g x+2 f y+c=0 \ldots .$. (i)

For point (3, -2),

Substituting x = 3 and y = -2 in (i), we get

9 + 4 + 6g – 4f + c = 0

⇒ 6g – 4f + c = -13 ….(ii)

For point (1, 0),

Substituting x = 1 andy = 0 in (i), we get

1 + 0 + 2g + 0 + c = 0

⇒ 2g + c = -1 ……(iii)

For point (-1, -2),

Substituting x = -1 and y = -2, we get

1 + 4 – 2g – 4f + c = 0

⇒ 2g + 4f – c = 5 …….(iv)

Adding (ii) and (iv), we get

8g = -8

⇒ g = -1

Substituting g = -1 in (iii), we get

-2 + c = -1

⇒ c = 1

Substituting g = -1 and c = 1 in (iv), we get

-2 + 4f – 1 = 5

⇒ 4f = 8

⇒ f = 2

Substituting g = -1, f = 2 and c = 1 in (i), we get

$x^2+y^2-2 x+4 y+1=0$

If (1, -4) satisfies equation (v), the four points are concyclic. Substituting x = 1, y = -4 in L.H.S of (v), we get

L.H.S. $=(1)^2+(-4)^2-2(1)+4(-4)+1$

= 1 + 16 – 2 – 16 + 1

= 0

= R.H.S.

Point (1, -4) satisfies equation (v).

∴ The given points are concyclic.