# How do you find the equation for the circle where A(7,-6),B(9,2) and C(1,4) are points on a circle?

Aug 6, 2016

${\left(x - 4\right)}^{2} + {\left(y + 1\right)}^{2} = 34$

#### Explanation:

Notice that $\vec{A B} = \left(2 , 8\right)$ and $\vec{B C} = \left(- 8 , 2\right)$ are perpendicular to one another and of the same length.

So $A B C$ is an isosceles right angled triangle. Together with the point $D \left(- 1 , - 4\right)$ the points $A , B , C , D$ form the vertices of a square with centre at the midpoint of $A C$:

$E = \left(\frac{7 + 1}{2} , \frac{- 6 + 4}{2}\right) = \left(4 , - 1\right)$

The radius of the circumscribing circle is the distance between the centre $E$ and any of $A , B , C , D$ say $C$:

$r = \sqrt{{\left(1 - 4\right)}^{2} + {\left(4 - \left(- 1\right)\right)}^{2}} = \sqrt{{3}^{2} + {5}^{2}} = \sqrt{34}$

The equation of a circle with centre $\left(h , k\right)$ and radius $r$ may be written:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

Hence the equation of our circle may be written:

${\left(x - 4\right)}^{2} + {\left(y - \left(- 1\right)\right)}^{2} = 34$

or slightly easier to read as:

${\left(x - 4\right)}^{2} + {\left(y + 1\right)}^{2} = 34$

graph{((x-4)^2+(y+1)^2-34)((x-4)^2+(y+1)^2-0.03)((x-7)^2+(y+6)^2-0.08)((x-9)^2+(y-2)^2-0.08)((x+1)^2+(y+4)^2-0.08)((x-1)^2+(y-4)^2-0.08) = 0 [-14.67, 25.33, -10.44, 9.56]}