How do you find the equation of a circle that passes through points (-8,-2)(1-,-11) and (-5,9)?

Question

1899 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-01T11:47:10

The equn. of a circle is :

$7x^2+7y^2-52x-10y-912=0$

Let the equation of the circle be

$x^2+y^2+2gx+2fy+c=0,$

Since the circle passes through $(-8,-2), (1,-11)$ and $(-5,9)$

$(-8)^2+(-2)^2-16g-4f+c=0$
$1^2+(-11)^2+2g-22f+c=0$
$(-5)^2+9^2-10g+18f+c=0$

These equations simplify to

$-16g-4f+c+68=0$ $-----------(1)$
$2g-22f+c+122=0$ $-----------(2)$
$-10g+18f+c+106=0$ $----------(3)$

Subtracting $(1)$ from $(2)$
$18g-18f+54=0$ $------------(4)$

Subtracting $(1)$ from $(3)$
$6g+22f+38=0$ $-------------(5)$

Multiply $(5)$ by 3
$18g+66f+114=0$ $------------(6)$

Subtracting $(4)$ from $(6)$
$84f+60=0$

$therefore f= -5/7$

Substituting the value of $f$ into $(4)$

$18g+90/7+54=0$
$18g+468/7=0$
$18g=-468/7$

Divide both sides by 18
$g=-26/7$

Substituting the value of f and g into $(1)$
$-16(-26/7)-4(-5/7)+c+68=0$
$416/7+20/7+c+68=0$
$c+912/7=0$
$c=-912/7$

Hence, the equation of the circle is

$x^2+y^2+2(-26/7)x+2(-5/7)y-912/7=0$
$x^2+y^2-52/7x-10/7y-912/7=0$

Multiply through by 7
$therefore 7x^2+7y^2-52x-10y-912=0$

In the form
$(x-a)^2+(y-b)^2=r^2$

$(x-26/7)^2+(y-5/7)^2=7085/49$