Answer the following :

Show that the points (9, 1), (7, 9), (−2, 12) and (6, 10) are concyclic

#### Solution

Let the equation of circle passing through the points (9, 1), (7, 9), (–2, 12) be

x^{2} + y^{2} + 2gx + 2fy + c = 0 …(i)

For point (9, 1),

Substituting x = 9 and y = 1 in (i), we get

81 + 1 + 18g + 2f + c = 0

∴ 18g – 2f + c = –82 …(ii)

For point (7, 9),

Substituting x = 7 and y = 9 in (i), we get

49 + 81 + 14g + 18f + c = 0

∴ 14g + 18f + c = – 130 …(iii)

For point (–2, 12),

Substituting x = – 2 and y = 12 in (i), we get

4 + 144 – 4g + 24f + c = 0

∴ –4g + 24f + c = – 148 …(iv)

By (ii) – (iii), we get

4g – 16f = 48

∴ g – 4f = 12 ,,,(v)

By (iii) – (iv), we get

18g – 6f = 18

∴ 3g – f = 3 ..(vi)

By 3 x (v) – (vi), we get

– 11f = 33

∴ f = – 3

Substituting f = – 3 in (vi), we get

3g – (– 3) = 3

∴ 3g + 3 = 3

∴ g = 0

Substituting g = 0 and f = – 3 in (ii), we get

18 (0) + 2(– 3) + c = – 82

∴ – 6 + c = – 82

∴ c = –76

∴ Equation of the circle becomes

x^{2} + y^{2} + 2(0)x + 2(– 3)y + (– 76) = 0

∴ x^{2} + y^{2} – 6y – 76 = 0 …(vii)

Now for the point (6, 10),

Substituting x = 6 and y = 10 in L.H.S. of (vii),

we get

L.H.S = 6^{2} + 10^{2} – 6(10) – 76

= 36 + 100 – 60 – 76

= 0

= R.H.S.

∴ Point (6,10) satisfies equation (vii).

∴ the given points are concyclic.