A circle passes through the points (1, 2), (3, 4), and (x1, y1). Its center is equal to (x1, y1) when the radius is equal to 0. Calculate the area of a circle of radius 25 units.

Steps to Solve

1. Identify the center and radius of the circle

The general equation of a circle in standard form is given by: $$ (x - h)^2 + (y - k)^2 = r^2 $$ where ((h, k)) is the center of the circle and (r) is the radius.

Based on the problem, the center point is given as ((1, 2)) and the radius is specified as (3).

2. Write the equation of the circle

Substituting the values for the center ((h, k) = (1, 2)) and radius (r = 3) into the general equation of the circle, we get: $$ (x - 1)^2 + (y - 2)^2 = 3^2 $$ This simplifies to: $$ (x - 1)^2 + (y - 2)^2 = 9 $$

3. Verify the intersection with the line

The line is given as (y - 3 = -\frac{4}{3}(x - 1)). Rearranging this to slope-intercept form gives: $$ y = -\frac{4}{3}x + \frac{4}{3} + 3 $$ Simplifying results in: $$ y = -\frac{4}{3}x + \frac{13}{3} $$

4. Substitute the line equation into the circle equation

Substituting (y) from the line equation into the circle's equation: $$ (x - 1)^2 + \left(-\frac{4}{3}x + \frac{13}{3} - 2\right)^2 = 9 $$ We need to solve this quadratic equation for (x).

5. Solve for (x)

Expanding and simplifying the left side will give us a quadratic equation in the form of: $$ ax^2 + bx + c = 0 $$

We can use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

6. Find the intersection points and verify the radius

After finding (x), use the corresponding (y) value from the line equation to determine the intersection points. This verifies the location of the points concerning the circle's radius.