Equation of a circle with radius and contains 2 points Find the equation of the circle with radius $5$ and contains the points $A(-8,0)$ and $B(-4,-2)$
Hint: Equate radii.
Thats the question from my midterms exam, can anyone help me finding this one, its a little tricky for me. Thanks, hope anyone can cite a good solution for me to follow. Thanks again..
 A: The centre of the circle lies on the perpendicular bisector of the line segment that joins $A$ and $B$. 
The midpoint of this line segment is $(-6,-1)$. The line segment has slope $-\frac{1}{2}$, so the perpendicular bisector has slope $2$.
Now we can find the equation of the perpendicular bisector. It is $y=2x+11$.
The distance from $(x,2x+11)$ to $(-8,0)$ is $5$. Thus
$$(x+8)^2+(2x+11)^2=25.$$
This simplifies to 
$$5x^2+60x+160=0,$$
and then to $x^2+12x+32=0$. The roots of this equation are easy to find: we get lucky, the quadratic even factors nicely. 
A: $$(8+x_1)^2 + y_1^2 = 5^2$$
$$(4+x_1)^2 + (2+y_1)^2 = 5^2$$
Subtracting the second equation from first:
$$4(12+2x_1) - 2(2+2y_1) = 0$$
$$11 + 2x_1 = y_1$$
Substitute this in one of the two equations and find $x_1$ and then eventually $y_1$.Once you get the center of the circle $(x_1,y_1)$, you can calculate the equation of the circle using:
$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
where $r=5$ in this case.
A: General eq. of circle is 
$(x-a)^2+(y-b)^2=r^2$  where r is radius & $(a,b)$ is the center.
Hint:Put value of $r=5$ then
$(x-a)^2+(y-b)^2=25$ 
Now put points which are situated on circle in place of (x,y) one by one you get two eq. solve them for values of $a$ & $b$ then you get the eq. of circle by using these values in general eq.
