Question

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

• A. $x^2+y^2=20$
• B. $x^2+y^2-2x-4y=0$
• C. ${(x}^2+y^2-4)+\lambda (x^2+y^2-16)=0$
• D. N.O.T.

Answer 1

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

Let $(x-h{)}^2+(y-k{)}^2=r^2$ …………..$\left(1\right)$

where (h, k) is the centre of the circle and r is the radius.

Let us draw a line $A(2,0)$ and $B(4,0)$. If the centre lies outside the line AB then it is a chord.

The diameter of the circle is bigger than the chord. Suppose AB is the diameter of the circle then the centre of the circle must be a midpoint of AB.

$\therefore$ Using section formula

$h=\frac{2+0}{2}=\frac{2}{2}=1$

$k=\frac{4+0}{2}=\frac{4}{2}=2$

$\therefore$ Coordinates of the centre$=(h,k)$

$=(1,2)$

Equation $\left(1\right)$ becomes,

$(x-1{)}^2+(y-2{)}^2=r^2$ …………….$\left(2\right)$

Since $\left(2\right)$ passes through $(2,0)$, equation $\left(2\right)$ can be written as,

$(2-1{)}^2+(0-2{)}^2=r^2$

$1^2+2^2=r^2$

$1+4=r^2$

$r^2=5$

$r=\sqrt{5}$

Equation of the circle with minimum radius is

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

$\Rightarrow x^2+1-2x+y^2+4-4y=5$

$\Rightarrow x^2+y^2-2x-4y+5-5=0$

$\Rightarrow x^2+y^2-2x-4y=0$.