The standard equation of a circle is
(x-h)^2 +(y-k)^2 = r^2
where
(h,k)is the center.
"The circle is tangent to the y-axis at y=3."
So a radius, r, is perpendicular to the y-axis at y=3.
Therefore, r = h, and k=3.
Draw the figure on paper.
Draw the horizontal radius h from (0,3) to (h,3) to see the relations.
Draw the radius h from (1,0) to (h,3).
By distance between two points,
h = sqrt[(1-h)^2 +(0-3)^2]
h^2 = (1 -2h +h^2) +9
2h = 10
h = 5
Draw the vertical radius to intersect the x-axis.
The horizontal chord, lying on the x-axis, of the circle is bisected by this vertical radius.
Since h=5, the intersection of this vertical radius and this horizontal chord is at (5,0).
Therefore, the (1,0) is 4 units to the left of (5,0).
Hence, the other end of the chord is at (9,0). <----the other x-intercept, answer.
Since (h,k) = (5,3), and r=h=5, the equation of the circle is
(x-5)^2 +(y-3)^2 = 25 ---------------answer.
-----------------
Check if the (9,0) satisfies the equation.
(9-5)^2 +(0-3)^2 =? 25
16 +9 =? 25
25 =? 25
Yes, so, OK.
Hello, konnie!
This is a variation of ticbol's solution.
. . Refer to Captain Black's diagram.
A circle is tangent to the y-axis at and has an x-intercept at .
(a) Determine the other x-intercept
(b) Deduce the equation of the circle.
The center of the circle is: .
The y-intercept is: .
The x-intercept is: .
We find that . . and .
Since and are radii: .
. . and we have:. .
Hence, the center is and
(b) Therefore: .
Let
. .
(a) The other x-intercept is: .