Exploring the Equation of a Circle with Given Normals and Tangents

In this article, we delve into the problem of finding the equation of a circle given that it has two normals and a tangent line. Specifically, the normals are defined by the equation xy - 2x - y^2 0 and a tangent line given by 3x 4y 6. The article will explore the mathematical derivations to find the equation of the circle that satisfies these given conditions.

Normalization of Normals

The given equation of the normals, xy - 2x - y^2 0, can be factored into two linear equations. To do this, we can rewrite the equation as:

xy - y^2 - 2x 0

This can be further factored into:

y(x - 1) - 2(x - 1) 0

Simplifying, we get:

(x - 1)(y - 2) 0

From this factorization, we see that the center of the circle is at the intersection of the lines x - 1 0 and y - 2 0. Thus, the center of the circle is located at the point (1, 2).

Equation of the Circle Family

A circle with an arbitrary center (x0, y0) and radius r has the general equation:

(x - x0)^2 (y - y0)^2 r^2

Given the center (1, 2), the equation of the family of circles is:

(x - 1)^2 (y - 2)^2 r^2

Where r is the radius of the circle.

Finding the Radius Using the Tangent Line

To find the specific radius that fits the given tangent line, we use the distance formula from a point to a line. The formula to find the distance d from a point (x0, y0) to the line Ax By C 0 is:

d |Ax0 By0 C| / sqrt(A^2 B^2)

In our case, the tangent line is 3x 4y - 6 0. The point (1, 2) is the center of the circle. Substituting into the distance formula, we get:

r |3(1) 4(2) - 6| / sqrt(3^2 4^2)

r |3 8 - 6| / sqrt(9 16)

r |5| / sqrt(25)

r 5 / 5

r 1

Thus, the radius of the circle is 1.

Final Equation of the Circle

Substituting the radius back into the family of circles equation, we get:

(x - 1)^2 (y - 2)^2 1

This is the equation of the circle that meets the given conditions of having the normals xy - 2x - y^2 0 and the tangent line 3x 4y 6.

Conclusion

In this exploration, we have derived the equation of a circle given its normals and a tangent line. The key steps involved factoring the normal equation to find the center of the circle, using the distance formula to find the radius, and then substituting back into the circle equation. This method provides a clear and systematic approach to solving similar problems in geometry and analytic geometry.