TITLE: Find the equation of the circle containing the point $(-2,2)$ and passing through..
QUESTION [0 upvotes]: Find the equation of the circle containing the point $(-2,2)$ and passing through the points of intersection of the two circles $C_1:\; x^2+y^2+3x-2y-4=0$ and  $C_2: x^2+y^2-2x-y-6=0$
In my attempt to solve this question, I tried solving the equation simultaneously. But the results don't seems like something I can work with $(x,y)=\{0.215, 3.075\} \{-0.715, -1.575\}$
Does anyone suggest I move on with these figures? Or. Is this a wrong start?

REPLY [1 votes]: You now have three noncolinear points that lie on the circle. I expect that you’ve learned how to find an equation for this circle from their coordinates. One can write the equation directly in the form of a determinant: $$\begin{vmatrix}x^2+y^2&x&y&1\\x_1^2+y_1^2&x_1&y_1&1\\x_2^2+y_2^2&x_2&y_2&1\\x_3^2+y_3^2&x_3&y_3&1\end{vmatrix} = 0.$$ Plug in the coordinates of the three points, expand the determinant, and simplify.  
However, you can find the equation of the circle without computing the intersection points explicitly. Every circle that passes through those points has an equation that’s an affine combination of the two given equations: $$(1-\lambda)(x^2+y^2+3x-2y-4)+\lambda(x^2+y^2-2x-y-6)=0.$$ You can plug in the coordinates $(-2,2)$ and solve for $\lambda$, or use Plücker’s mu to obtain an equation directly: Given the intersection $S$ of the solution sets of $f(x,y)=0$ and $g(x,y)=0$, and a point $P=(x_0,y_0)$ not in $S$, an equation with a solution set that contains both $S$ and $P$ is $f(x_0,y_0)g(x,y)-g(x_0,y_0)f(x,y)=0$. Applying this to the two circles, we get $$-6(x^2+y^2-2x-y-6)-4(x^2+y^2+3x-2y-4) = 0,$$ which I’ll leave to you to simplify.
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