Find the Equation of a Circle Tangent to a Line at a Given Point

In this article, we will explore the process of finding the equation of a circle that touches a given line at a specific point and whose center lies on another line. This problem requires a blend of algebraic manipulation and geometric principles. Let's break it down step by step.

Problem Statement

We are given the equation of a line as 3x - 4y - 35 0 and a point of tangency A(5,5). We need to find the equation of the circle whose center lies on the line x - y - 1 0 and is tangent to the given line at the point A(5,5).

Step-by-Step Solution

Step 1: Determine the Center of the Circle

Let's denote the center of the circle as (h, k). Since the center lies on the line x - y - 1 0, we can express h in terms of k:

h - k - 1 0 implies h k 1

Step 2: Determine the Radius of the Circle

The radius r of the circle is the distance from the center (h, k) to the point A(5,5). Using the distance formula, we get:

r sqrt{(h - 5)^2 (k - 5)^2}

Step 3: Apply the Tangency Condition

For the circle to be tangent to the line 3x - 4y - 35 0, the distance from the center (h, k) to the line must equal the radius r. The distance d from a point (x_0, y_0) to the line Ax By C 0 is given by:

d frac{|Ax_0 By_0 C|}{sqrt{A^2 B^2}}

For our line 3x - 4y - 35 0, we have A 3, B -4, and C -35. Thus the distance from (h, k) to the line is:

d frac{|3h - 4k - 35|}{sqrt{3^2 (-4)^2}} frac{|3h - 4k - 35|}{5}

Setting this distance equal to the radius, we get:

frac{|3h - 4k - 35|}{5} sqrt{(h - 5)^2 (k - 5)^2}

Step 4: Substitute h k 1 into the Equation

Substituting h k 1 into the distance equation, we get:

frac{|3(k 1) - 4k - 35|}{5} sqrt{(k 1 - 5)^2 (k - 5)^2}

This simplifies to:

frac{|3k 3 - 4k - 35|}{5} sqrt{(k - 4)^2 (k - 5)^2}

frac{|-k - 32|}{5} sqrt{(k - 4)^2 (k - 5)^2}

The left side further simplifies to:

frac{32 - k}{5} (since -k - 32 - (k 32))

Step 5: Simplify the Right Side

The right side simplifies to:

sqrt{(k - 4)^2 (k - 5)^2} sqrt{k^2 - 8k 16 k^2 - 10k 25} sqrt{2k^2 - 18k 41}

Step 6: Set the Equations Equal and Solve for k

Equating both sides, we have:

frac{32 - k}{5} sqrt{2k^2 - 18k 41}

Multiplying both sides by 5, we get:

32 - k 5sqrt{2k^2 - 18k 41}

Squaring both sides, we get:

(32 - k)^2 25(2k^2 - 18k 41)

Expanding both sides, we have:

1024 - 64k k^2 50k^2 - 900k 1025

Rearranging terms, we get:

49k^2 - 836k 1 0

This quadratic equation can be solved using the quadratic formula:

k frac{836 pm sqrt{836^2 - 4 cdot 49 cdot 1}}{2 cdot 49}

Solving this, we find:

k 1

Step 7: Determine the Center of the Circle

Using h k 1, we find:

h 1 1 2

Therefore, the center of the circle is (2, 1).

Step 8: Determine the Radius

The radius r is the distance from the center (2, 1) to the point A(5,5):

r sqrt{(2 - 5)^2 (1 - 5)^2} sqrt{(-3)^2 (-4)^2} sqrt{9 16} 5

Step 9: Write the Equation of the Circle

The equation of the circle is:

(x - 2)^2 (y - 1)^2 25

Final Answer

The equation of the circle is:

boxed{(x - 2)^2 (y - 1)^2 25}