Question

Find the equation of the circle passing through the points (1,2) and(2,1) and centre is on y-axis

Answer 1

Answer:

Required Equation of Circle is   $x^{2} +y^2=\sqrt{5}$.

Step-by-step explanation:

Let the centre of the circle = ( 0,k )

We know that,

Equation of the circle = $(x-0)^2+(y-k)^2=r^2$

Now, the equation of the circle passing through the points ( 1,2 ) and ( 2,1 )

Putting the Point ( 1,2 ) in equation of Circle

⇒  $(1-0)^2+(2-K)^2=r^2$

⇒ $1+4+k^2-4k=r^2$

⇒ $k^2-4k+5=r^2$             equ. (1)

Putting the Point ( 2,1 ) in equation of Circle

⇒ $(2-0)^2+(1-K)^2=r^2$

⇒ $4+1+k^2-2k=r^2$

⇒ $k^2-2k+5=r^2$            equ. (2)

From equ. (1) and equ. (2), we have

K = 0

Putting value of k in equ. (1)

r = $\sqrt{5}$

Hence, Required Equation of Circle is:

$x^{2} +y^2=\sqrt{5}$