Question

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

Answer 1

See below

Explanation:

Circle general equation is

`(x-a)^2+(y-b)^2=r^2` where `(a,b)` 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 `abs(-5)=5`

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

Answer 2

I assume the notation `C(-5,6)` means the center is at `(x_c,y_c)=(-5,6)`

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 `(0,6)`)

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

I will use
`color(white)("XXX")(x-x_c)^2+(y-y_c)^2=r^2`
as the "standard form" for a circle with center `(x_c,y_c)` and radius `r`.

So, in this case we have
`color(white)("XXX")(x+5)^2+(y-6)^2=5^2`

...or, maybe your version of the standard form might be an expansion of this:
`color(white)("XXX")x^2+10x+y^2-12y+36=0`