Yo sup??
The given points are (1,1),(1,3) and (9,2)
equation of circle is
8x²+8y²+ax+by=0
let us plug in (1,1) in the equation we get
8+8+a+b=0
a+b=-16
let us now plug in (1,3) in the equation
8+8*3²+a+3b=0
a+3b=-80
let us. now solve the two equation in terms of a and b
we get
2b=--64
b=-32
and
a=-16+32
=16
therefore the equation is
8x²+8y²+16x-32y=0
Hope this helps
Answer:
[tex] \rm \displaystyle 8{x}^{2} + 8{y}^{2} \underline{- 79}x \underline{ - 32}y \underline{ + 95 }= 0[/tex]
Step-by-step explanation:
we are given 3 points which lie in a circle of the circle equation we want to figure out missing constant and coefficients
our given equation
[tex] \rm \displaystyle 8{x}^{2} + 8{y}^{2} + Ax + By + C = 0[/tex]
to figure out the missing coefficients and the constant we can consider system of equation because we have three points by using the points we can create the system of equations from the first point we obtain:
[tex] \rm \displaystyle A + B+ C = - 16[/tex]
from the second point we acquire:
[tex] \rm \displaystyle 8{.1}^{2} + 8{.3}^{2} + A.1+ B.3+ C = 0[/tex]
simplify:
[tex] \rm \displaystyle A+ 3B+ C = - 80[/tex]
from the third point we get:
[tex] \rm \displaystyle 8{.9}^{2} + 8{.2}^{2} + A.9 + B.2 + C = 0[/tex]
simplify:
[tex] \rm \displaystyle 9 A+ 2B + C= - 680[/tex]
therefore our system of equations is:
[tex] \begin{cases}A + B + C = - 16 \\A + 3B + C = - 80 \\9A + 2B + C = - 680\end{cases}[/tex]
by solving the equation we acquire:
[tex] \displaystyle \rm A = - 79,B = - 32,C = 95[/tex]
substitute hence, our equation is
[tex] \rm \displaystyle 8{x}^{2} + 8{y}^{2} - 79x - 32y + 95 = 0[/tex]
and we are done!
