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

May 2, 2018

A circle of radius 1, centred at the point $\left(2 , 4\right)$

#### Explanation:

${\left(y - 4\right)}^{2} + {\left(x - 2\right)}^{2} = 1$

Note: a circle of radius $r$ centred at the point $\left(a , b\right)$ has the equation: ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

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

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

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

Jul 7, 2018

See below:

#### Explanation:

The equation of a circle is given by

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

With center $\left(h , k\right)$ and radius $r$.

From our equation

${\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} = 1$

We know that our center is at $\left(2 , 4\right)$ 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!