Question

The equation of the smallest circle passing through the points $(2,2)$ and $(3,3)$ is

• A. $x^2+y^2+5x+5y+12=0$
• B. $x^2+y^2-5x-5y+12=0$
• C. $x^2+y^2+5x-5y+12=0$
• D. $x^2+y^2-5x+5y+12=0$

Answer 1

The correct option is B $x^2+y^2-5x-5y+12=0$

If the smallest circle is drawn through the points $(2,2)$ and $(3,3)$ then, the point have to be the opposite end of the diameter of the circle.
Therefore, the centre of the circle is $C=(\frac{5}{2},\frac{5}{2})$.
The radius of the circle is $R=\frac{\sqrt{2}}{2}$ ....(using distance formula)
Hence equation of the circle is
$(x-\frac{5}{2}{)}^2+(y-\frac{5}{2}{)}^2=\frac{1}{2}$
$x^2+y^2-5x-5y+\frac{50}{4}=\frac{1}{2}$
Or
$x^2+y^2-5x-5y+\frac{25}{2}-\frac{1}{2}=0$
Or
$x^2+y^2-5x-5y+12=0$.