Pre-Calculus (show complete solution)

Question

Find the standard form of the equation of the circle, if the center of the circle of radius 3 lies on the line 𝑥+2𝑦=4 and it is tangent to the y – axis.

Answer 1

The standard form of a circle is (x - h)² + (y - k)² = r², where (h,k) is the center and r is the radius.

Since x + 2y = 4, 2y = 4-x. So, y = (4-x) / 2

The center has the form (x, (4-x)/2).

Since the circle is tangent to the y-axis, x = 3 or -3

So, the center is (3, 1/2) or (-3, 7/2)

There are 2 possibilities for the equation: (x - 3)² + (y - 1/2)² = 9 or (x+ 3)² + (y - 7/2)² = 9

Answer 2

There are actually 2 different circle equations that would answer this question. Each circle is the same size, is tangent to the y-axis, and has a center on the line x + 2y = 4, but one is on the left side and one is on the right side of the y-axis.

This the generic standard equation for any circle, but we will need to update it for our specific circles.

(x - h)² + (y - k)² = r²

The x and y stand for x- and y- coordinates of any point on the edge of the circle, and these letters should not be replaced with numbers in our final answer.

The h and k stand for the x- and y- coordinates of the center of the circle, and we should replace these letters with numbers in our final answer, once we figure out what the center of each circle is.

The r stands for the radius of the circle, and we should replace this with 3 in our final answer because 3 is the radius of these circles. Now our equation is:

(x - h)² + (y - k)² = 3²

Squaring the 3, we get:

(x - h)² + (y - k)² = 9

Now, let's figure out the coordinates of the center of the circles. If the circles are tangent to the y-axis, the y-axis touches the edge of the circles. The x-coordinate of the y-axis is 0. Because the radius of the circle is 3, the center of the circle must be 3 units away from the y-axis. Thus, the x-coordinate of the center of the circle is either at 3 or -3. Both options would produce a circle with a radius of 3 that is tangent to the y-axis; they'd just be on opposite sides of the y-axis.

Now that we know the x-coordinates of the center of the circles (the h's), we can plug them into the equation of the line that the center of the circle lies on, in order to find the y-coordinates of the circles (the k's).

For the circle on the left side of the y-axis, x = -3:

x + 2y = 4 <-- original equation

-3 + 2y = 4 <-- replaced x with -3

2y = 7 <-- added 3 to both sides

y = 3.5 <-- divided both sides by 2

For the circle on the right side of the y-axis, x = 3:

x + 2y = 4 <-- original equation

3 + 2y = 4 <-- replaced x with 3

2y = 1 <-- subtracted 3 from both sides

y = 0.5 <-- divided both sides by 2

We now have everything we need for standard equations for the circles.

Generic Equation: (x - h)² + (y - k)² = r²

Equation for circle #1: (x + 3)² + (y - 3.5)² = 9

Equation for circle #2: (x - 3)² + (y - 0.5)² = 9

Here's a visual: https://www.desmos.com/calculator/tj5le6kz5k

I hope that helps! :^)