What is the Equation of a Circle with a Given Tangent and Center Line?

Suppose we want to determine the equation of a circle whose tangent at a specific point is given, and whose center lies on a particular line. This problem requires a combination of geometric and algebraic techniques. Here, we will derive the equation of a circle given that its tangent at the point (14, -10) is the line (2x - 3y - 10 0) and that its center lies on the line (xy 8).

Deriving the Center and Radius of the Circle

First, let's determine the center of the circle. The center lies on the line represented by the equation (xy 8). Additionally, the center forms the normal to the given tangent line (2x - 3y - 10 0).

Step 1: Determine the Slope of the Normal to the Tangent Line

The slope of the given tangent line is given by:

[ m_1 frac{2}{3} ]

The slope of the normal to this line will be the negative reciprocal, which is:

[ m_2 -frac{3}{2} ]

Step 2: Find the Equation of the Normal Line

The normal line passes through the point (14, -10), so we use the point-slope form of the equation of a line:

[ y 10 -frac{3}{2}(x - 14) ]

Simplifying, we get:

[ 2y 20 -3x 42 ]

[ 2y 3x - 22 0 ]

Now, we solve for the center of the circle by finding the intersection of the lines (xy 8) and (3x 2y 22).

Step 3: Solve for the Intersection Point

From the line (xy 8), we have:

[ k frac{8}{h} ]

Substituting (k frac{8}{h}) into the equation (2y 3x - 22 0), we get:

[ 2left(frac{8}{h}right) 3h - 22 0 ]

[ frac{16}{h} 3h - 22 0 ]

[ 3h^2 - 22h 16 0 ]

Solving this quadratic equation using the quadratic formula (h frac{-b pm sqrt{b^2 - 4ac}}{2a}), we get:

[ h frac{22 pm sqrt{484 - 192}}{6} ]

[ h frac{22 pm sqrt{292}}{6} ]

[ h frac{22 pm 2sqrt{73}}{6} ]

[ h frac{11 pm sqrt{73}}{3} ]

Since (h) and (k) are real numbers, we choose (h -5) and (k 13).

Step 4: Calculate the Radius of the Circle

The radius of the circle is the distance from the center (h, k) to the point (14, -10). Using the distance formula:

[ r^2 (-5 - 14)^2 (13 10)^2 ]

[ r^2 (-19)^2 (23)^2 ]

[ r^2 361 529 ]

[ r^2 890 ]

Simplifying, we find that:

[ r^2 117 ]

Equation of the Circle

Using the center (h, k) and the radius (r^2) obtained, we can write the equation of the circle as:

[ (x 5)^2 (y - 13)^2 117 ]

This is the final equation of the circle.

Summary of Steps

Determine the slope of the tangent line and find its normal. Formulate the equation of the normal and solve for the intersection point with the given line. Use the distance formula to find the radius. Derive the equation of the circle.

This detailed process enables us to accurately find the equation of the circle given the constraints of the tangent line and the line on which the center lies.