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

(a) Determine the other x-intercept

(b) Deduce the equation of the circle.

My answer: EH? please help!

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- Aug 12th 2007, 10:38 PMkonnieHelp with a seemingly simple math problem!
A circle is tangent to the y-axis at y=3 and has one x-intercept at x=1.

(a) Determine the other x-intercept

(b) Deduce the equation of the circle.

My answer: EH? please help! - Aug 12th 2007, 11:29 PMCaptainBlack
- Aug 13th 2007, 01:57 AMticbol
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. - Aug 13th 2007, 05:58 AMSoroban
Hello, konnie!

This is a variation of ticbol's solution.

. . Refer to Captain Black's diagram.

Quote:

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

(a) Determine the other x-intercept

(b) Deduce the equation of the circle.

The center of the circle is: .$\displaystyle C(h,3)$

The y-intercept is: .$\displaystyle A(0,3)$

The x-intercept is: .$\displaystyle B(1,0)$

We find that .$\displaystyle CA\,=\,h$ . and .$\displaystyle CB \:=\:\sqrt{(h-1)^2 + 3^2}$

Since $\displaystyle CA$ and $\displaystyle CB$ are radii: .$\displaystyle h \:=\:\sqrt{(h-1)^2 + 9}$

. . and we have:. . $\displaystyle h^2 \:=\;h^2 - 2h + 1 + 9\quad\Rightarrow\quad 2h \:=\:10\quad\Rightarrow\quad h \,=\,5$

Hence, the center is $\displaystyle C(5,3)$ and $\displaystyle r\,=\,5$

**(b)**Therefore: .$\displaystyle \boxed{{\color{blue}(x-5)^2 + (y-3)^2 \:=\:25}}$

Let $\displaystyle y = 0\!:\;\;(x-5)^2 + (0-3)^2 \:=\:25\quad\Rightarrow\quad (x-5)^2 + 9 \:=\:25$

. . $\displaystyle (x-5)^2\:=\:16\quad\Rightarrow\quad x-5 \:=\:\pm4\quad\Rightarrow\quad x \:=\:1,\,9$

**(a)**The other x-intercept is: .$\displaystyle \boxed{{\color{blue}(9,\,0)}}$

- Aug 14th 2007, 11:01 PMkonnie
Thank you ^___^