Question

The equation of the circle which passes through (2,0)
and (0,5) and its radius as small as possible is​

Answer 1

We can draw infinitely many circles passing two points. When does the radius become smallest? It is when the two points are on the diameter.

According to the fact

The radius is least where (2,0) and (0,5) are the two endpoints of the diameter.

The least radius is $\dfrac{\sqrt{29} }{2}$.

The equation is $(x-1)^2+(y-\dfrac{5}{2} )^2=\dfrac{29}{4}$.

There are infinitely many circles passing two points!

Let's find the circle. A circle has its center and radius. Assume $(a,b)$ and r.

$\implies (x-a)^2+(y-b)^2=r^2$

(2,0) and (0,5) lies on the graph of the circle.

$\implies\displaystyle{\left \{ {{(2-a)^2+b^2=r^2} \atop {a^2+(5-b)^2=r^2}} \right. }$

We get two equations. If we cancel out $r^2$ by subtraction, we get the following equation.

This line is which the radius of the circle $(a,b)$ draws.

$\implies 4a-10b-21=0$