Calc Problem: find the equation of this circle.?

Question

A circle is tangent to the y axis at y=3 and has an x intercept at x=1.

a. Determine the other x intercept

b. Deduce the equation of the circle

Answer 1

i'll first find the equation of the circle , i don't think you would neccessaily need calculus to do it

so first step would be to draw a diagram

so one point of circle would be (0,3) which is point of tangency let this be point A

another point on circle would be (1,0) which is point of intercept with x-axis let this be point B

since the the circle is tangent at point (0,3 )then the diameter of circle (parallel to x-axis ) would be (x, 3) let this be point C

with these 3 points draw a triangle

the slope of the line AB would be -3

and notice that since the triangle is inscribed within the lower part of the circle, based on Thales' theorem, this triangle must be a right triangle

so the line BC would have a slope that is negative reciprocal of AB, 1/3

now with the slope calculate the value of x

1/3 = 3/ (x-1)

x = 10

so the radius of the circle would be (10-0)/2 = 5

and the centre of circle (h,k) is (3,5)

now put it into form of (x-h)^2 + (y-k)^2 = r^2

and equation of circle (x-5)^2 + (y-3)^2 = 25

for the x intercept substitute y with zero and solve for x

(x-5)^2 + (-3)^2 = 25

(x-5)^2 = 16

(x-5) = 4 or x-5 = -4

x= 9 x= 1

since the one of them is already x= 1

then the other x-intercept is x= 9

hope this helps

Answer 2

The first answerer missed a couple of valid points that shouldn't be glazed over:

First, you need to point out the definition of a circle being the set of all points equadistant from a given point called its center, say (Cx,Cy) as the center.

Then you need to discuss that a tangent to a circle is always perpendicular to a line drwn from the point of tangency to the center.  In other words the point about the center being (x,3) - I know what the answerer meant, but only because I already knew how to do the problem.  

Really, the way to say it is since the tangent line is the y axis, a segment from the point of tangency at (0,3) to the circle's center would be perpendicular to the y-axis, or parallel to the x-axis with a slope of 0.  Now we know more about the center - instead of Cx,Cy we can now call it (Cx,3) knowing that the y value of the center must be 3 to make its slope 0.  Cy = 3.

Now you take (Cx,3) and use the distance formula on it with the two points you know on the circle of (0,3) and (1,0).  Obvoiusly the distance between (Cx,3) and (0,3) is the value of Cx.  This distance between the center and any point on circle is known as the radius.

General equation of circle: [(x-Cx)^2]+[(y-Cy)^2]= radius^2

That being said, plug in what you know.  You know radius is Cx and you know two points satisfying the equation:

for (1,0):

(1-Cx)^2+(0-3)^2=Cx^2.

When you multiply out the first term, you are able to "cancel out the Cx^2 term on both sides, leaving you with Cx=5.

Now you know the center is (5,3) and the radius is 5.  Use the general formula:

(x-5)^2+(y-3)^2=25.

The first answerer's plan only works because all the numbers were whole and easily drawn on a piece of graph paper.  You can't always "just see" the answer.  

You have to know how to do it!