How to Find the Equation of a Circle Given Its Touch and Intercept Conditions

When dealing with circles in advanced mathematics, one common task is to derive their equation given specific conditions. This article will explore how to determine the equation of a circle that touches the x-axis at a specified distance and intercepts a specified length on the y-axis. We will walk through the steps using detailed examples and mathematical derivations.

Understanding the Problem

The problem at hand involves finding the equation of a circle that:

Touches the x-axis at a distance of 3 from the origin. Intercepts a length of 6 on the y-axis.

Let's delve into the mathematical details of how to solve this problem step-by-step. This example will cover multiple scenarios to arrive at all possible circles that meet these conditions.

Step-by-Step Solution

Step 1: Identifying the Centers of the Circles

Consider the centers of the circles as points A(4, h) and B(-4, -h). Here, h is the height of the center from the x-axis. The circles will share a common chord CD of length 6. The midpoint of CD, denoted as M, is at a distance of 3 from D.

Step 2: Using the Right Triangle Property

Consider the right triangle ADM, where AM is the radius of the circle, DM is half the length of the chord, and AD is the distance from the center to the x-axis. Applying the Pythagorean theorem:

[AM^2 - DM^2 AD^2]

[4^2 - 3^2 h^2]

[16 - 9 h^2]

[h^2 7]

[h 5]

Therefore, the height from the x-axis to the center of the circles is 5 units.

Step 3: Formulating the Circle Equations

With the height known, the equations of the circles can be derived. The general form of the circle equation is:

[(x - h)^2 (y - k)^2 r^2]

For the circle centered at 4, 5 and -4, -5, the equations are:

[(x - 4)^2 (y - 5)^2 25]

[(x 4)^2 (y 5)^2 25]

These represent the circles in the first, fourth, second, and third quadrants, respectively.

Step 4: Verifying and Deriving the Final Equations

Let's derive the final equations by verifying the problem conditions.

1st Possibility: Circle in the First Quadrant

The circle touches the positive x-axis at (4, 0) and cuts off an intercept BC 6 units on the y-axis. Let O(4, 5) be the center and r be the radius. Draw a perpendicular OD on BC from O, thus BD CD 3 units.

In the right-angled triangle ODB:

[r^2 3^2 4^2]

[r^2 9 16]

[r^2 25]

[r 5]

Center O(4, 5). The equation of the circle is:

[(x - 4)^2 (y - 5)^2 25]

[x^2 - 8x 16 y^2 - 10y 25 25]

[x^2 y^2 - 8x - 10y 16 0]

2nd Possibility: Circle in the Second Quadrant

The circle touches the negative x-axis at (-4, 0), with the same BC 6 units on the y-axis. Let O(-4, -5) be the center and r 5. The equation of the circle is:

[(x 4)^2 (y 5)^2 25]

[x^2 8x 16 y^2 10y 25 25]

[x^2 y^2 8x 10y 16 0]

3rd Possibility: Circle in the Third Quadrant

The circle is centered at (-4, -5). The equation is:

[(x 4)^2 (y 5)^2 25]

[x^2 8x 16 y^2 10y 25 25]

[x^2 y^2 8x 10y 16 0]

4th Possibility: Circle in the Fourth Quadrant

The circle is centered at (4, -5). The equation is:

[(x - 4)^2 (y 5)^2 25]

[x^2 - 8x 16 y^2 10y 25 25]

[x^2 y^2 - 8x 10y 16 0]

Conclusion

In summary, the equations of the circles that meet the given conditions are:

[x^2 y^2 - 8x - 10y 16 0] [x^2 y^2 8x - 10y 16 0] [x^2 y^2 8x 10y 16 0] [x^2 y^2 - 8x 10y 16 0]

These equations represent the circles in the four quadrants, all satisfying the given conditions.