Question

The circle passing through the distinct points (1, t), (t, 1) and (tt) for all values of t, passes through the point

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

Q1

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Q2

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Q3

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Q4

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Q5

 

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Q6

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Q7

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Q8

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Q9

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Q10

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