<s>[INST] Equation of circle with minimum radius that can be drawn through the points $A$ and $B.$ The line $y=mx+c$ cuts the given circle $x^2+y^2=a^2$ at two distinct points $A$ and $B.$ Equation of circle having minimum radius that can be drawn through the points $A$ and $B.$
I have approached the problem this way:
Considering it intersects at two points $A$ and $B.$
Now solving equation of circle and line.
So we get an equation in terms of $x$ and $y$
$$x^2(1 + m^2) + c^2 + 2mcx = a^2$$
and
$$y^2(1 + m^2) -2cy + c^2 - a^2m^2 = 0$$
These are the equations of $x$ coordinate and $y$ coordinate of two points $A$ and $B.$
Now since minimum radius circle can be drawn only when $AB$ is the diameter of the circle, the centre of the circle is the midpoint of $AB,$ which comes out to be $\left(\frac{-mc}{1+m^2} , \frac{c}{1+m^2}\right)$
and the radius is half of the distance $|AB|.$
But this way equation of circle would be difficult to obtain.
So I am interested in any other way of approaching.
 [/INST] A circle passing through $A$ and $B$ would have an equation $x^2+y^2-a^2+k(mx-y+c)=0$ (where $k$ is a constant) and its centre is located at $\displaystyle \left(\frac{-km}{2},\frac{k}{2}\right)$. The circle with the minimum radius will have $AB$ as a diameter and its centre lies on $AB$. 
So, we have $\displaystyle m\left(\frac{-km}{2}\right)-\left(\frac{k}{2}\right)+c=0$. This implies that $\displaystyle k=\frac{2c}{m^2+1}$.
The equation of the required circle is  $\displaystyle x^2+y^2-a^2+\frac{2c}{m^2+1}(mx-y+c)=0$, i.e. $\displaystyle x^2+y^2+\frac{2mc}{m^2+1}x-\frac{2c}{m^2+1}y-a^2+\frac{2c^2}{m^2+1}=0$
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