Question

Write the equation in standard form of a circle with center (-3,6), tangent to the y-axis.

Answer 1

Answer:

$(x+3)^2+(y-6)^2=9$

Step-by-step explanation:

The equation of a circle in standard form is written as

$(x-a)^2+(y-b)^2=r^2$

where

(a, b) are the coordinates of the center

r is the radius of the circle

In this problem, the circle is centered at (-3,6), so we have

a = -3

b = 6

Also, we know that the circle is tangent to the y-axis, and the x-coordinate of its center is 3: this means that the horizontal distance between the point of the circle tangent to the y-axis and the centre of the circle is 3, therefore the radius of the circle is 3.

So, the equation of this circle is:

$(x-(-3))^2+(y-6)^2=3^2$

$(x+3)^2+(y-6)^2=9$