Question

The equation of a circle which passes through $(1,2)$ and $(2,1)$ and whose radius is $1unit$ is

• A. $x^2+y^2-4x-4y+7=0$
• B. $x^2+y^2-2x-2y+1=0$
• C. $x^2+y^2-4x-2y+7=0$
• D. $x^2+y^2-x-y+7=0$

Answer 1

Answer: (A), (B)

The correct options are
A $x^2+y^2-4x-4y+7=0$
B $x^2+y^2-2x-2y+1=0$

General equation of a circle is

${(x-h)}^2+{(y-k)}^2=r^2$

Given: The circle passes through the points $(1,2)$ and $(2,1)$ and radius is $1$ units.So, we will put the value of $(x,y)=(1,2)$ and $(2,1)$

${(1-h)}^2+{(2-k)}^2=1^2$ ......$\left(1\right)$

${(2-h)}^2+{(1-k)}^2=1^2$ ......$\left(2\right)$

$\Rightarrow {(1-h)}^2+{(2-k)}^2={(2-h)}^2+{(1-k)}^2$ from $\left(1\right)$ and $\left(2\right)$

$\Rightarrow 1+h^2-2h+4+k^2-4k=4+h^2-4h+1+k^2-2k$

$\Rightarrow -2h-4k=-4h-2k$

$\Rightarrow -2h-4k+4h+2k=0$

$\Rightarrow 2h-2k=0$

$\therefore h=k$

Put the value of $h$ in $\left(1\right)$ we get

${(1-h)}^2+{(2-h)}^2=1^2$

$\Rightarrow 1+h^2-2h+4+h^2-4h=1$

$\Rightarrow 2h^2-6h+4=0$

$\Rightarrow h^2-3h+2=0$

$\Rightarrow h^2-2h-h+2=0$

$\Rightarrow h(h-2)-1(h-2)=0$

$\Rightarrow (h-2)(h-1)=0$

$\therefore h=1,2$

$\Rightarrow k=1,2$ since $h=k$

For $(h,k)=(1,1)$

${(x-1)}^2+{(y-1)}^2=1$ is the general equation of a circle

or $x^2+1-2x+y^2+1-2y=1$

or $x^2+y^2-2x-2y+1=0$

For $(h,k)=(2,2)$

${(x-2)}^2+{(y-2)}^2=1$ is the general equation of a circle

or $x^2+4-4x+y^2+4-4y=1$

or $x^2+y^2-4x-4y+7=0$

Hence the equations $x^2+y^2-2x-2y+1=0$ and $x^2+y^2-4x-4y+7=0$ represent the required equation of circle.