How do I write an equation for a circle in standard given only 3 points: (2,3), (6,-7), and (-4,9)?

Question

267 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-21T12:53:00

The equation of circle is
#x^2+y^2+41.11x+24.94y -157.06 =0#

Let the equation of circle be #x^2+y^2+2gx+2fy +c= 0#

Points #(2,3), (6,-7),(-4,9)# will satisfy the equation.

#:. 2^2+3^2+2g*2+2f*3 +c= 0 # or

# 13+4g +6f+c= 0 or 4g +6f+c= -13(1) #, similarly

#:. 6^2+(-7)^2+2g*6-2f*7 +c= 0 # or

# 85+12g -14f+c= 0 or 12g -14f+c= -85(2) #, similarly

#:. (-4)^2+9^2-2g*4+2f*9 +c= 0 # or

# 97-8g +18f+c= 0 or -8g +18f+c= -97(3) #,

Solving equation(1) , equation(2) and equation(3) we get

#g~~20.556, f~~12.472, c ~~ -157.056# . Hence the equation

of circle is #x^2+y^2+41.11x+24.94y -157.06 = 0#

graph{x^2+y^2+41.11x+24.94y-157.06=0 [-80, 80, -40, 40]}

[Ans]