What is the standard form of the equation of a circle given points: (7,-1), (11,-5), (3,-5)?

1 Answer
Shwetank Mauria
Jun 21, 2017

Standard form of circle is (x−7)2+(y+5)2=16

Explanation:

Let the equation of circle be x2+y2+2gx+2fy+c=0, whose centre is (−g,−f) and radius is √g2+f2−c. As it passes though (7,−1), (11,−5) and (3,−5), we have

49+1+14g−2f+c=0 or 14g−2f+c+50=0 ......(1)

121+25+22g−10f+c=0 or 22g−10f+c+146=0 ...(2)

9+25+6g−10f+c=0 or 6g−10f+c+34=0 ......(3)

Subtracting (1) from (2) we get

8g−8f+96=0 or g−f=−12 ......(A)

and subtracting (3) from (2) we get

16g+112=0 i.e. g=−7

putting this in (A), we have f=−7+12=5

and putting values of g and f in (3)

6×(−7)−10×5+c+34=0 i.e. −42−50+c+34=0 i.e. c=58

andequation of circle is x2+y2−14x+10y+58=0

and its centre is (7,−5) abd radius is √49+25−58=√16=4

and standard form of circle is (x−7)2+(y+5)2=16

graph{x^2+y^2-14x+10y+58=0 [-3.08, 16.92, -9.6, 0.4]}

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