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Show that the Points (3, −2), (1, 0), (−1, −2) and (1, −4) Are Concyclic.

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Question

Show that the points (3, −2), (1, 0), (−1, −2) and (1, −4) are concyclic.

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Solution

Let the required equation of the circle be

\[x^2 + y^2 + 2gx + 2fy + c = 0\]...(1)
It is given that the circle passes through (3, −2), (1, 0), (−1, −2).
∴ \[13 + 6g - 4f + c = 0\]    ...(2)
\[1 + 2g + c = 0\] ...(3)
\[5 - 2g - 4f + c = 0\] ...(4)
Solving (2), (3) and (4):
\[g = - 1, f = 2, c = 1\] 
Theerefore, the equation of the circle is
\[x^2 + y^2 - 2x + 4y + 1 = 0\] ...(5)
We see that the point (1, −4) satisfies the equation (5).
Hence, the points (3, −2), (1, 0), (−1, −2) and (1, −4) are concyclic.
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Advanced Concept of Circle - Standard Equation of a Circle
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Chapter 24: The circle - Exercise 24.2 [Page 32]

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R.D. Sharma Mathematics [English] Class 11
Chapter 24 The circle
Exercise 24.2 | Q 5 | Page 32

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