Question

WILL GIVE BRAINLIEST!

Answer 1

Answer:

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

Step-by-step explanation:

General Equation Of A Circle

If a,b, and c are real numbers, then the general equation of a circle is

$\displaystyle x^2+y^2+ax+by+c=0$

There are three unknowns that could eventually be determined if we knew three points of the circle. These points are (1, 7) (8, 6) (7, -1). We only need to replace them in the general equation and solve the resulting system of equations

For the point (1, 7)

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

Operating and rearranging

$\displaystyle a+7b+c=-50$

For the point (8, 6)

$\displaystyle (8)^2+(6)^2+a(8)+b(6)+c=0$

Operating and rearranging

$\displaystyle 8a+6b+c=-100$

For the point (7, -1)

$\displaystyle (7)^2+(-1)^2+a(7)+b(-1)+c=0$

Operating and rearranging

$\displaystyle 7a-b+c=-50$

We form the system of equations

$\left\{\begin{matrix}a+7b+c=50\\ 8a+6b+c=-100\\ 7a-b+c=-50\end{matrix}\right.$

We'll eliminate c from the first two equations and then from the last two equations

Multiplying the first one by -1

$\left\{\begin{matrix}-a-7b-c=50\\ 8a+6b+c=-100\end{matrix}\right.$

Adding them up we have

$\displaystyle 7a-b=-50$

Adding up with the third equation

$\displaystyle 7a-b+c=-50$

We get

c=0

Knowing this value, let's return to the original system with c=0

$\left\{\begin{matrix}a+7b=-50\\ 8a+6b=-100\end{matrix}\right.$

Multiplying the first by -8

$\left\{\begin{matrix}-8a-56b=400\\ 8a+6b=-100\end{matrix}\right.$

Adding them up

$\displaystyle -50b=300$

Which gives  

$b=-6$

Finally, isolating a from

$a=-50-7b$

We get

a=-8

So the general equation of the circle is

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