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four distinct point (2k,3k),(1,0),(0,1) and (0,0) lies on a circle for k equal to
Asked by vs.super.king999 | 28 Apr, 2024, 09:37: AM

General equation of circle is

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

If the circle as given in eqn.(1)  pases through (1,0) , the we get

2g+c = -1 ...................(2)

If the circle as given in eqn.(1)  pases through (0,1) , the we get

2f+c = -1 ...................(3)

from eqn.(2) and (3) , we get

g = f ..........................(4)

If the circle as given in eqn.(1)  pases through (0,0) , the we get

c = 0 ...................(5)

Using eqn.(5) , we get from equations (2) , (3) and (4) as

g = f = -1/2

Hence equation of circle is

x2 + y2  - x - y  = 0  ............................(6)

If the circle as given in eqn.(6)  pases through (2k,3k) , the we get

4k2 + 9k2 - 2k - 3k = 0

13k2 - 5 k = 0

hence we get k = 5/13

Answered by Thiyagarajan K | 28 Apr, 2024, 10:07: AM

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