Question

The equation of the circle passing through $(2,0)$ and $(0,4)$ and having the minimum radius is

• A. $x^2+y^2-4$
• B. $x^2+y^2-2x+4y=0$
• C. $x^2+y^2-x-2y=0$
• D. $x^2+y^2-2x-4y=0$

Answer 1

The correct option is D $x^2+y^2-2x-4y=0$

The circle passing through the given two points have minimum radius when those points are diametric ends.

The equation of cirlce with center (h,k) and radius r is $(x-h{)}^2+(y-k{)}^2=r^2$

The circle with minimum radius will have,

Center = $(\frac{2+0}{2},\frac{0+4}{2})=(1,2)$

Radius = $\frac{\sqrt{(2-0{)}^2+(0-4{)}^2}}{2}=\frac{\sqrt{20}}{2}=\sqrt{5}$

So, equation of circle,

$\to (x-1{)}^2+(y-2{)}^2=5$

$\to x^2+y^2-2x-4y=0$

Hence, (D) is the correct answer.