Question

Show that the points (3, −2), (1, 0), (−1, −2) and (1, −4) are concyclic

Answer 1

Let the equation of the circle passing through

the points (3, – 2), (1, 0) and (– 1, – 2) be

x² + y² + 2gx + 2fy + c = 0 …(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 and y = 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

x² + y² – 2x + 4y + 1 = 0 …(v)

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.