Question

50 points and BRAINLIEST!

Answer 1

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:

$(x-x_{c})^{2} +(y-y_{c})^{2}=r^{2}$

where $r$ is the radius, whilst $x_{c}$ and $y_{c}$ 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:

$x^{2} +y^{2}+ Dx+Ey+F=0$    Eqn (1).

where $D$, $E$ and $F$ are constant coefficients that can be solved by plugging in the points given (x,y) and creating a System of Linear Equations as follow:

• Point (1,7) in Eqn (1):

        $1^2 +7^2 + D + 7E+F=0$,

        $D+7E+F=-50$   Eqn (2)  

• Point (8,6) in Eqn (1):

       $8^2+6^2+8D+6E+F=0$

        $8D+6E+F=-100$ Eqn (3)  

• Point (7,-1) in Eqn (1):

        $7^2+(-1)^2+7D-E+F=0$

        $7D-E+F=-50$      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:

$Eqns (1)-(2): \\D+7E+F+50 - [8D+6E+F+100] =\\D+7E+F+50-8D-6E-F-100=\\-7D+E-50$

which solving for E we have $E=50+7D$   Eqn. (5)

Lets do the same and subtract Eqn(4) from (2) so that:

$Eqn (2)-(4):\\D+7E+F+50-[7D-E+F+50]=\\D+7E+F+50-7D+E-F-50=\\-6D+8E=0$

which solving for E we have $E=\frac{6D}{8}$  Eqn. (6)

Equating Eqns (5) and (6) we then obtain:

$\frac{6D}{8}=50+7D\\6D=8(50+7D)\\6D=400+56D\\6D-56D=400\\D=-8$

and thus plugging in D in Eqn. (6) we have:

$E=\frac{6(-8)}{8}\\E=-6$

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)):

$-8+7(-6)+F=-50\\-8-42+F=-50\\-50+F=-50\\F=-50+50\\F=0$

So finally we have that $D=-8, E=-6, F=0$ so our equation of the circle will eventually read:

$x^2+y^2-8x-6y=0$

Answer 2

Answer:

$x^2+y^2-8x-6y=0$.

Step-by-step explanation:

Let the general equation of the circle be $x^2+y^2+2ax+2by+c=0$.

Each of the points must satisfy this equation.

We substitute (1,7) to get:

$1^2+7^2+2a*1+2b*7+c=0$.

$50+2a+14b+c=0$.

$2a+14b+c=-50....(1)$.

We substitute (8,6) to get:

$8^2+6^2+2a*8+2b*6+c=0$.

$100+16a+12b+c=0$.

$16a+12b+c=-100....(2)$.

We substitute (7,-1) to get:

$7^2+(-1)^2+2a*7+2b*(-1)+c=0$.

$50+14a-2bb+c=0$.

$14a-2b+c=-50....(3)$.

Solving equations 1, 2, and 3 simultaneously gives:$a=-4,b=-3,c=0$

Hence the equation of the circle is

$x^2+y^2+2*-4*x+2*-3*y+0=0$.

$x^2+y^2-8x-6y=0$.