Find the circle passing through (-1, -3) and (-5, 3) and having its center on the line x-2y+2=0. Please answer if you can... this question is quite difficult for me. Thanks! ^_^
 A: HINT:
If $(a,b)$ be the coordinate of the center, $a-2b+2=0\implies a=2b-2\ \ \ \ (1)$
Now, $\displaystyle(a+1)^2+(b+3)^2=(a+5)^2+(b-3)^2=$the square of the radius 
$\displaystyle\implies 2a+1+6b+9=10a+25-6b+9\implies 8a-12b+24=0\ \ \ \ (2)$
Solve for $a,b$ and immediately you have the radius $\displaystyle\sqrt{(a+1)^2+(b+3)^2}$
A: Hint:
Assume the center is $(x,y)$. Equate the distances from the center to your two points to get one equation and as center lies on the line you have the points eligible to be the circle's center. Once you have the center you can get the radius which is the distance between any of the point to the center
A: Correct me if I'm wrong, but this question can be answered without calculus. The center of the circle must be of the form $(x_0,x_0/2+1)$. Thus the circle will an equation of the form $(x-x_0)^2+(y-x_0/2-1)^2=r^2$. We know $(-1,-3)$ and $(-5,3)$ are solutions. So we have
$$(x_0+1)^2+(x_0/2+4)^2=r^2$$
$$(x_0+5)^2+(x_0/2-2)^2=r^2$$
Simplifying,
$$5x_0^2/4+6x_0+17-r^2=0$$
$$5x_0^2/4+8x_0+29-r^2=0$$
Subtracting these equations gives $x_0=-6$, and correspondingly $y_0=x_0/2+1=-2$. So the equation of the circle is
$$(x+6)^2+(y+2)^2=r^2$$
$r$ can be found be calculating the distance from the center to either of the given points. We finally obtain
$$(x+6)^2+(y+2)^2=26$$
A: suppose that center has coordinate $(x,y)$ ,then clearly this equation satisfy   line equation namely
$x-2*y+2=0$
also if it is passing through two points,then distance  till both  point are equal,because we have
$(x-a)^2+(y-b)^2=R^2$
we would have
$(x+1)^2+(y+3)^2=(x+5)^2+(y-3)^2$
after several  algebraic equation we would get
$2*x+1+6*y+9=10*x+25-6*y+9$
or
$8*x-12*y=0$ and 
$x-2*y+2=0$
please could you solve this?
