What is the equation of a circle that is tangent to the y-axis where y equals 3 and has an x-intercept of 1 and another x-intercept?

Answer 1

Since the line x=0 is tangent to the circle at (0,3), the center of the circle must lie on the line y=3 (because it is the only line perpendicular to x=0 and containing (0,3)). Since the circle must also contain point (1,0), the center must be on the perpendicular bisector of (1,0) and (0,3). The midpoint of (1,0) and (0,3) is (1/2,3/2)(using the midpoint formula). The slope of the line containing (0,3) and (1,0) is -3/1(using the slope formula). Since we are looking for a line perpendicular to the line containing (0,1) and (3,0), the slope of the line must be the opposite of the reciprocal of -3/1, which is 1/3. Since the slope is 1/3 and it must contain the point (1/2,3/2), one equation of the line(in point-slope form) must be y-3/2=1/3(x-1/2). Since the center of the circle must lie on both of these lines, we can solve the following system of equations:

y=3

y-3/2=1/3(x-1/2)

Since y is given by the first equation, we can substitute it in the second equation. This gives 3-3/2=1/3(x-1/2). By simplifying, this is equivalent to 3/2=x/3-1/6, which, by adding 1/6 to both sides, is equivalent to (18+12)/12=x/3, which, by simplifying, is equivalent to 20/12=x/3. By multiplying both sides by 3, we get 20/4=x, which is equivalent to 5=x.

Since x=5 and y=3, the center must be at (5,3). The distance from (5,3) to (0,3) is 5 units since the y coordinate is the same and 5-0=5.

Now, since we have the radius and the center, we can put these values into the equation for a circle, giving (y-3)^2+(x-5)^2=5^2. Since 5^2=25, the equation is equivalent to (y-3)^2+(x-5)^2=25.