Answer:
The equation of the circle passing through the points: (1, 7) (8, 6) (7, -1) is x^2+y^2-8x-6y=0
Step-by-step explanation:
In principle there are multiple ways one can find the equation of a circle given 3 points. For this case, we shall use the most common one, i.e. the General Equation of a Circle. In theory the equation actually reads as:
[tex](x-x_{c})^{2} +(y-y_{c})^{2}=r^{2}[/tex]
where [tex]r[/tex] is the radius, whilst [tex]x_{c}[/tex] and [tex]y_{c}[/tex] are points on the circle. Now since here the information given is not as such we shall use the alternate equation of the Circle passing through points, given as:
[tex]x^{2} +y^{2}+ Dx+Ey+F=0[/tex] Eqn (1).
where [tex]D[/tex], [tex]E[/tex] and [tex]F[/tex] are constant coefficients that can be solved by plugging in the points given (x,y) and creating a System of Linear Equations as follow:
[tex]1^2 +7^2 + D + 7E+F=0[/tex],
[tex]D+7E+F=-50[/tex] Eqn (2)
[tex]8^2+6^2+8D+6E+F=0[/tex]
[tex]8D+6E+F=-100[/tex] Eqn (3)
[tex]7^2+(-1)^2+7D-E+F=0[/tex]
[tex]7D-E+F=-50[/tex] Eqn (4)
Now we have a system of three linear equations (i.e. Eqns (2), (3) and (4)) that can be solved to find the constants D, E and F as follow.
Lets start by subtracting Eqn (3) from (2) so that:
[tex]Eqns (1)-(2): \\D+7E+F+50 - [8D+6E+F+100] =\\D+7E+F+50-8D-6E-F-100=\\-7D+E-50[/tex]
which solving for E we have [tex]E=50+7D[/tex] Eqn. (5)
Lets do the same and subtract Eqn(4) from (2) so that:
[tex]Eqn (2)-(4):\\D+7E+F+50-[7D-E+F+50]=\\D+7E+F+50-7D+E-F-50=\\-6D+8E=0[/tex]
which solving for E we have [tex]E=\frac{6D}{8}[/tex] Eqn. (6)
Equating Eqns (5) and (6) we then obtain:
[tex]\frac{6D}{8}=50+7D\\6D=8(50+7D)\\6D=400+56D\\6D-56D=400\\D=-8[/tex]
and thus plugging in D in Eqn. (6) we have:
[tex]E=\frac{6(-8)}{8}\\E=-6[/tex]
Finally we can find F using our D and E values and any of the Eqns (2),(3) or (4) as follow (using Eqn (1)):
[tex]-8+7(-6)+F=-50\\-8-42+F=-50\\-50+F=-50\\F=-50+50\\F=0\\[/tex]
So finally we have that [tex]D=-8, E=-6, F=0[/tex] so our equation of the circle will eventually read:
[tex]x^2+y^2-8x-6y=0[/tex]
