Question

Please help with math problem

Answer 1

Answer:

$a_0 = -8$

$a_1=-6$

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

Step-by-step explanation:

The equation of the circle has the following form

$x^2 + y^2 + a_0x + a_1y=0$

The equation of the circle has the following form

We know that the circle goes through the following points

(1, 7)

(8, 6)

(7, -1).

Then we substitute the values of x and y in the equation

For (1, 7)

$(1)^2 + (7)^2 + a_0(1) + a_1(7)=0$

$1 + 49 + a_0 + 7a_1=0$

$a_0 + 7a_1=-50$    (1)

For  (8, 6)

$(8)^2 + (6)^2 + a_0(8) + a_1(6)=0$

$64 + 36 + 8a_0 + 6a_1=0$

$8a_0 + 6a_1= -100$    (2)

With these equations is enough to solve the system

$a_0 + 7a_1=-50$    (1)

$8a_0 + 6a_1= -100$    (2)

Multiply the equation (1) by -8 and add it to the equation (2)

$-8a_0 - 56a_1=400$    (1)

$8a_0 + 6a_1= -100$    (2)

-----------------------------------------------------

$-50a_1 = 300\\\\a_1 = -6$

Then

$a_0 + 7(-6) = -50\\\\a_0 = -50+42\\\\a_0=-8$

Finally the equation is:

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