Answer:
(x-24/5)^2 + (y-21/5)^2 = (sqrt(481)/5)^2, or (x-24/5)^2 + (y-21/5)^2 = 481/25
Explanation:
The equation of a circle can be written in the form (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center of the circle and r is its radius. Since the lines are tangent to the circle at points (0,5) and (6,5), we can use these points to find the center and radius of the circle.
First, let’s find the slope of each line. The slope of the line y = 3/4x+5 is 3/4. The slope of the line 3x + 4y = 38 can be found by rearranging it into slope-intercept form: y = (-3/4)x + 19/2. So the slope of this line is -3/4.
Since the lines are tangent to the circle, they are perpendicular to the radius at their point of tangency. This means that the center of the circle lies on the line that passes through (0,5) and has a slope of -4/3, and also on the line that passes through (6,5) and has a slope of 4/3.
Let’s find the equation of these two lines. The line passing through (0,5) with a slope of -4/3 has an equation y - 5 = (-4/3)(x - 0), or y = (-4/3)x + 5. The line passing through (6,5) with a slope of 4/3 has an equation y - 5 = (4/3)(x - 6), or y = (4/3)x - 3.
The center of the circle is at the intersection of these two lines. To find it, we can set their right-hand sides equal to each other and solve for x: (-4/3)x + 5 = (4/3)x - 3. Solving this equation gives us x = 24/5. Substituting this value into either equation for y gives us y = (4/3)(24/5) - 3 = 21/5.
So the center of the circle is at (24/5, 21/5). To find its radius, we can use either point of tangency. Let’s use (0,5). The distance between this point and the center is given by sqrt((0-24/5)^2 + (5-21/5)^2), which simplifies to sqrt(481)/5.
Therefore, the equation of the circle is (x-24/5)^2 + (y-21/5)^2 = (sqrt(481)/5)^2, or (x-24/5)^2 + (y-21/5)^2 = 481/25.