Question

Prove that the given points are concyclic (9, 1), (7, 9), (-2, -12), (6, 10).

Answer with a unique method would be highly appreciable.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that centre of circle be (h, k) and radius of circle be r units.

So, equation of circle is

$\rm \: {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2} - - - (1)$

As it is given that circle passes through the point (9, 1).

$\rm \: {(9 - h)}^{2} + {(1 - k)}^{2} = {r}^{2}$

$\rm \: 81 + {h}^{2} - 18h + {k}^{2} + 1 - 2k = {r}^{2}$

$\rm \: {h}^{2} + {k}^{2} - 18h - 2k + 82 = {r}^{2} - - - (2)$

Also, circle (1) passes through (7, 9)

$\rm \: {(7 - h)}^{2} + {(9 - k)}^{2} = {r}^{2}$

$\rm \: 49 + {h}^{2} - 14h + {k}^{2} + 81 - 18k = {r}^{2}$

$\rm \: {h}^{2} + {k}^{2} - 14h - 18k + 130= {r}^{2} - - - (3)$

Also, given that circle (1) passes through (6, 10)

$\rm \: {(6 - h)}^{2} + {(10 - k)}^{2} = {r}^{2}$

$\rm \: 36 + {h}^{2} - 12h + 100 + {k}^{2} - 20k = {r}^{2}$

$\rm \: {h}^{2} + {k}^{2} - 12h - 20k + 136 = {r}^{2} - - - (4)$

On equating equation (2) and (3), we get

$\rm \: {h}^{2} + {k}^{2} - 18h - 2k + 82 = {h}^{2} + {k}^{2} - 14h - 18k + 130$

$\rm \: - 18h - 2k + 82 = - 14h - 18k + 130$

$\rm \: - 4h + 16k = 48$

$\rm \: 4( - h + 4k) = 48$

$\rm \: - h + 4k = 12 - - - - (5)$

On equating equation (3) and (4), we get

$\rm \: {h}^{2} + {k}^{2} - 14h - 18k + 130 = {h}^{2} + {k}^{2} - 12h - 20k + 136$

$\rm \: - 14h - 18k + 130 = - 12h - 20k + 136$

$\rm \: - 2h + 2k = 6$

On dividing both sides by 2, we get

$\rm \: - h + k = 3 - - - (6)$

On adding equation (5) and (6), we get

$\rm \: 5k = 15$

$\rm\implies \:k \: = \: 3$

On substituting k = 3 in equation (6), we get

$\rm\implies \:h \: = \: 0$

On substituting the values of h and k in equation (2), we get

$\rm \: 0 + 9 - 6 + 82 = {r}^{2}$

$\rm \: 85 = {r}^{2}$

$\rm\implies \: {r}^{2} = 85$

On substituting the values of h, k and r in equation (1), we get

$\rm \: {x}^{2} + {(y - 3)}^{2} = 85 - - - (7)$

Now, we have to justify whether (-2, - 12) lies on this circle or not.

Substituting (- 2, - 12,) in equation (7), we get

$\rm \: {( - 2)}^{2} + {( - 12 - 3)}^{2} = 85$

$\rm \: 4 + 225 = 85$

$\rm \: 229 = 85$

$\rm \: \: which \: is \: not \: possible$

Hence, points (9, 1), (7, 9), (-2, -12), (6, 10) are not concylic.

Answer 2

Correct Question - Prove that the given points are concyclic (9, 1), (7, 9), (-2, 12), (6, 10).

Assume -

g = h

f = k