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

$x^2+y^2-5x-5y+12=0$

Form the equation of the circle passing through the given points

The smallest circle passing through the points $(2,2)$ and $(3,3)$ must have these points diametrically opposite to each other

Hence the centre of the circle is the midpoint of these two points.

Let the centre of the circle be $C$ and radius of the circle be $R$

$C=(\frac{2+3}{2},\frac{2+3}{2})$

$C=\left(\frac{5}{2},\frac{5}{2}\right)...\left(i\right)$

By distance formula

$R=\sqrt{{\left(2-\frac{5}{2}\right)}^2+{\left(2-\frac{5}{2}\right)}^2}$

$R=\frac{1}{\sqrt{2}}...\left(ii\right)$

From $\left(i\right),\left(ii\right)$ the equation of the circle can be written as

${\left(x-\frac{5}{2}\right)}^2+{\left(y-\frac{5}{2}\right)}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$

$x^2+y^2-5x-5y+\frac{25}{2}=\frac{1}{2}$

$\Rightarrow$ $x^2+y^2-5x-5y+12=0$

Hence, option $\left(B\right)$ is the correct answer.