## Question

### Solution

Correct option is

(1, 1)

Let equation of the circle passing through the given points be

x2 + y2 + 2gx + 2fy + c = 0                  (1)

Then                   1 + t2 + 2g + 2ft + c = 0                      (2)

t2 + 1+ 2gt + 2f + c = 0                       (3)

and                     t2 + t2 + 2gt + 2ft + c = 0                     (4)

From (2) and (3), we have That is, f = g, because t ≠ 1, otherwise the three given points would be the same. Again, from (3) and (4), we get

t2 + 1+ 2gt + 2g + c = 0                                (5)

and                     2t2 + 2g t + 2 g t + c = 0                               (6)

Subtracting these, we get Finally, from (6), we get c = –2t2 + 2 t(t + 1) = 2t, so that the equation of the circle (1) can be written as

x2 + y2 – (t + 1)x – (t + 1)y + 2t = 0

which passes through (1, 1) for all values of t, while the points given by (b), (c) and (d) do not satisfy this equation.

#### SIMILAR QUESTIONS

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