Question

Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).

Answer 1

Answer:

$x^2+y^2-9.875x-4y+11.875=0$

Step-by-step explanation:

The general equation of the circle is

$x^2+y^2+ax+by+c=0$.

Putting in the point $(1,1)$, we get:

$1+1+a+b+c=0$

or

$\boxed{a+b+c=-2}$.

For the point $(1,3 )$ we get:

$1+9+a+3b+c =0$

or

$\boxed{a+3b+c =-10}$

Finally, for the point $(9,2 )$ we get:

$81+4+9a+2b+c =0$

or

$\boxed{ 9a+2b+c =-85}$

Thus, we have three equations and three unknowns:

$a+b+c=-2$.

$a+3b+c =-10$

$9a+2b+c =-85$

and their solutions are (Used Cramer's rule )

$a =-9.875$

$b = -4$

$c =11.875$

Thus, the equation of the circle is

$\boxed{ x^2+y^2-9.875x-4y+11.875=0}$