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

2 Answers
May 21, 2018

#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]}

May 21, 2018

#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)#

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