I encountered the following question while studying the chapter circles.
Find the equation of the circle passing through the points of intersection of the circles $x^2+y^2−2x−4y−4=0$ and $x^2+y^2−10x−12y+40=0$ and whose radius is 4.
I have figured out the following solution.
By using family of circles we get: $$C_1+\lambda C_2=0$$ After substituting the equations of circle, I get an equation in x, y and $\lambda$. I then simplified the equation in form of standard equation and got the radius in terms of $\lambda$. After equating it to the radius given in question, I got 2 values of $\lambda$. I then substituted $\lambda$ in the original equation to get two equations of circle.
Now, this process might seem easy but it is way too long and complicated. Is there any other shorter and a bit less complicated method to solve it?
It is not too hard to convince yourself by geometrical considerations that, since the line joining the centers of the two circles is the perpendicular bisector of the chord connecting the two intersection points, and also since the center of the circle(s) has to be distance 4 away from BOTH intersection points, it follows that it belongs to the perpendicular bisector, none other than the line joining the two centers!
Center of circle 1 is $P_1(1,2)$ and circle 2 $P_2(5,6)$ and thus the line joining them has equation $$y=2+\frac{6-2}{5-1}(x-1)=x+1$$
The equation of the chord is also not too hard to find, as pointed out by @Hyperion, by equating the two equations of the circle
$$y=\frac{44}{8}-x$$
This line is indeed perpendicular to the line joining the centers.
Now find the point of intersection of these two lines, which is the middle $M$ of the shared chord.
$$M:x+1=\frac{44}{8}-x\Rightarrow M(x_M, y_M=x_M+1)=M(\frac{9}{4}, \frac{13}{4})$$
Last thing we need to do is figure out the length of the chord, $2h$, where $h$ is the height of either triangle formed by $P_1, P_2$ and one of the intersection points. Since we know the distance between the centers $d=4\sqrt{2}$ and the radii of the circles $r_1=3, r_2=\sqrt{21}$, it's not too hard to show that the height is given by the formula
$$h^2=r_1^2-\frac{(r_1^2-r_2^2+d^2)^2}{4d^2}$$
and then the distance of the center of the circle of radius $4$ (we call it say $P(x_0, x_0+1)$) is is located on the line joining the centers, a distance from M equal to
$$|PM|=\sqrt{4^2-h^2}$$
and then the following quadratic equation has to be solved:
$$(x_0-x_M)^2+(x_0+1-y_M)^2=|PM|^2\Rightarrow 2(x_0-x_M)^2=|PM|^2\Rightarrow \{x_0=\frac{9}{4}\pm\frac{|PM|}{\sqrt{2}},y_0=\frac{13}{4}\pm\frac{|PM|}{\sqrt{2}}\}$$
and this gives the position of the center of the two circles possible.
In this problem $h^2=\frac{47}{8}$ and $|PM|=\frac{9}{2\sqrt{2}}$ which means that the two centers have coordinates
$$(0,1)~~,~~(9/2, 11/2)$$
In conclusion, I do not see this way being a lot simpler than the original, but it has the merit that there's no need to solve "complicated" quadratic equations.