a If the circle is tangent to the y-axis, it intersects the axis at one point.
B
b If the circle is tangent to the y-axis, it intersects the axis at one point.
A
a (x-2)^2+(y-6)^2=4
B
b (x-3)^2+(y-9)^2=9
Practice makes perfect
a The general equation of a circle is written in the following format.
CIRCLE
(x- a)^2+(y- b)^2= r^2
[-1em]
&Center: ( a, b)
&Radius: r
We already know that the circle's radius is 2. Let's substitute this into general equation of the circle.
From the exercise, we also know that the circle is tangent to the y-axis. This means it intersects the y-axis in only one point. Since the circle's radius is 2, the distance from the y-axis to the circle's center is 2. Also, because the circle is in the 1^(st) quadrant, the center's x-coordinate must be at x=2.
As we can see, the x-coordinate of our center is x=2. To calculate the corresponding y-coordinate, we have to substitute x=2 into the linear function and simplify.
Next, let's illustrate the circle in a diagram. This time, it will be 3 units to the right of the y-axis. Again, since the circle is in the 1^(st) quadrant, the midpoint's x-coordinate must be x=3.
To calculate the y-coordinate we have to substitute x=3 into the function and simplify.
The midpoint's y-coordinate is 9, which means the midpoint's coordinates are ( 3, 9). Now we can complete the equation of the circle.
(x- 3)^2+(y- 9)^2=9
