# How do you write the standard equation of a circle with the given center (-4,3) and tangent to the line y=1?

Jun 4, 2016

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

#### Explanation:

The center is already given, so the remaining item to obtain is the radius

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

where $\left(h , k\right)$ is the center of the circle

$\implies {\left(x + 4\right)}^{2} + \left(y - 3\right) = {r}^{2}$

The circle is tangent to the line $y = 1$.
We can get the radius by obtaining the distance between the center and point of tangency.

The point of tangency is at $\left(- 4 , 1\right)$.

Get the point of tangency by getting the intersection of the line passing through both the center and the point of tangency, and the tangent line.

Remember that the these lines are perpendicular. So the slope of one line should be equal to the negative inverse of the other.

Since the tangent line is $y = 1$, its slope is $0$. The slope of the perpendicular line is undefined, the equation line should be something like

$x = b '$

Since this line should pass through the center $\left(- 4 , 3\right)$, the equation of the line should be

$x = - 4$

Getting the intersection should yield us $\left(- 4 , 1\right)$

Now lets get the distance between the center and the point of tangency. This distance should be equal to the radius

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

=> D = sqrt((-4 - -4)^2 + (1 - 3)^2

$\implies D = \sqrt{{\left(- 2\right)}^{2}}$

$\implies D = 2$

$\implies r = 2$