How do you write an equation of a circle that passes through the points (3,6), (-1,-2), and (6,5)?

2 Answers

#x^2+y^2+4x-12y-25=0#

Explanation:

#x^2+y^2+2gx+2fy+c=0#

#9+36+6g+12f+c=0#
#6g+12f+c+45=0.....1#
#1+4-2g-4f+c=0#
#-2g-4f+c+5=0.....2#
#36+25+12g+10f+c=0#
#12g+10f+c+61=0....3#

by solving we get g=2, f=-6 c=-25
therefore the equation is #x^2+y^2+4x-12y-25=0#

Apr 5, 2018

#x^2+y^2-6*x-2*y-15=0#

Explanation:

This approach requires solving a system of three simultaneous first-degree equations.

Let the equation of the circle in a #x,y# plane be

#x^2+y^2+a*x+b*y+c=0#

where #a#, #b#, and #c# are unknowns.
Construct three equations about #a#, #b#, and #c#, one for each point given:
#3^2+6^2+3*a+6*b+c=0#,
#(1)^2+(-2)^2+(-1)*a+(-2)*b+c=0#, and
#6^2+5^2+6*a+5*b+c=0#

Solving for the system shall give
#a=-6#, #b=-2#, and #c=-15#

Thus the Equation of the circle:
#x^2+y^2-6*x-2*y-15=0#

Reference:
"[The] Equation of [a] circle passing through 3 given points", Maths Department, Queen's College, http://www.qc.edu.hk/math/Advanced%20Level/circle%20given%203%20points.htm