How do you graph #(y-4)^2+ (x-2)^2=1#?

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Answer 1 · 2018-05-02T09:22:51

A circle of radius 1, centred at the point #(2,4)#

#(y-4)^2 + (x-2)^2 =1#

Note: a circle of radius #r# centred at the point #(a,b)# has the equation: #(x-a)^2 + (y-b)^2 =r^2#

In this example: #a=2, b=2, r=1#

So, our graph is a circle of radius 1, centred at the point #(2,4)# as shown below:

graph{(y-4)^2 + (x-2)^2 =1 [-3.04, 9.444, -0.925, 5.315]}

Answer 2 · 2018-07-07T15:35:56

See below:

The equation of a circle is given by

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

With center #(h,k)# and radius #r#.

From our equation

#(x-2)^2+(y-4)^2=1#

We know that our center is at #(2,4)# and we have a radius of #1#. Now we can graph!

graph{(x-2)^2+(y-4)^2=1 [-2.924, 7.076, 1.44, 6.44]}

Hope this helps!