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Exercise 6.2 · Q24

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

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Let the circle through (3,−2)(3,-2), (1,0)(1,0), (−1,−2)(-1,-2) be x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0. Substituting (3,−2)(3,-2): 13+6g−4f+c=013+6g-4f+c=0 ...(A). Substituting (1,0)(1,0): 1+2g+c=0⇒c=−1−2g1+2g+c=0 \Rightarrow c=-1-2g ...(B). Substituting (−1,−2)(-1,-2): 5−2g−4f+c=05-2g-4f+c=0 ...(C). Putting (B) into (A): 12+4g−4f=0⇒g−f=−312+4g-4f=0 \Rightarrow g-f=-3 ...(D). Putting (B) into (C): 4−4g−4f=0⇒g+f=14-4g-4f=0 \Rightarrow g+f=1 ...(E). Adding (D) and (E) after solving: from (D) f=g+3f=g+3; substituting into (E): g+g+3=1⇒g=−1g+g+3=1 \Rightarrow g=-1, so f=2f=2, and c=−1−2(−1)=1c=-1-2(-1)=1. The circle through the …

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