Circle — Maths STD 11 Science — Question

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

For point (9, 1),

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

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

⇒ 18g + 2f + c = -82 …..(ii)

For point (7, 9),

Substituting x = 7 andy = 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 × (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

$\begin{aligned} & x^2+y^2+2(0) x+2(-3) y+(-76)= \\ & \Rightarrow x^2+y^2-6 y-76=0 \ldots \ldots(\text { vii) }\end{aligned}$

Now for the point (6, 10),

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

$\begin{aligned} & \text { L.H.S }=6^2+10^2-6(10)-76 \\ & =36+100-60-76 \\ & =0 \\ & =\text { R.H.S. }\end{aligned}$

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

∴ the given points are concyciic.