# Find the standard equation of the circle: C(-5,6) tangeant to y axis?

Jun 18, 2018

See below

#### Explanation:

Circle general equation is

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ where $\left(a , b\right)$ is center and $r$ is radius

If circle must be tangent to y axis, the distance from center to y axis to the given point is $\left\mid - 5 \right\mid = 5$

The equation is ${\left(x + 5\right)}^{2} + {\left(y - 6\right)}^{2} = 25$ graph{(x+5)^2+(y-6)^2=25 [-21.38, 18.62, -4.8, 15.2]}

Jun 18, 2018

I assume the notation $C \left(- 5 , 6\right)$ means the center is at $\left({x}_{c} , {y}_{c}\right) = \left(- 5 , 6\right)$

Since the circle is given as tangent to the y-axis
the distance from the center to the y-axis (i.e. the radius) is $5$
(since all points on the y-axis have $x = 0$, the tangent point will be $\left(0 , 6\right)$)

The required result may depend upon your understanding of what is meant by "the standard equation" for a circle.

I will use
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {r}^{2}$
as the "standard form" for a circle with center $\left({x}_{c} , {y}_{c}\right)$ and radius $r$.

So, in this case we have
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 5\right)}^{2} + {\left(y - 6\right)}^{2} = {5}^{2}$

...or, maybe your version of the standard form might be an expansion of this:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 10 x + {y}^{2} - 12 y + 36 = 0$