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!

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