Question

The equation of a circle and its graph passing through (1,1) and (1,-1) centred at (0-2) is?

Answer 1

`x^2+y^2+4y-6=0`

Explanation:

the general eqn of a circle centre `(a,b)` and radius`" "r`

`(x-a)^2+(y-b)^2=r^2`

we have centre (0,-2)#

`:. x^2+(y+2)^2=r^2`

we now have to find the radius

#the circle passes through

`(1,1)&(1,-1)`

substituting `(1,1)`

1+3^3=r^2

`r^2=10`

`:. x^2+(y+2)^2=10`

#expanding the eqn and simplifying

`x^2+y^2+4y+4=10`

`x^2+y^2+4y-6=0`

graph{x^2+y^2+4y-6=0 [-10, 10, -5, 5]}

Answer 2

`x^2+(y+2)^2=10`

Explanation:

Given -
Centre of the circle `(0,-2)`

We shall take point `(1,1)`

And Try to get the equation

State the formula when the circle's centre is not at the origin.

It is

`(x-h)^2+(y-k)^2=r^2`

Where -

`h=0` x coordinat of the circle
`k=-2`y coordinate of the circle

Let us rewrite the equation substituting the value of `x,y`

`(x-0)^2+(y+2)^2=r^2` ------------(1)

To find the equation, we need the value of `r`

To find that value, plug the value `(1,1)` in the equation (1)
because this point is on the circle. So, the equation must satisfy.

`(1-0)^2+(1+2)^2=r^2`
`1^2+3^2=r^2`
`1+9=r^2`
`r^2=10`

The required equation is -

`x^2+(y+2)^2=10`

The other point `(1, -1)` is on this cirlce.
I think the value is incorrect.
The correct value must be `(3, -1)`