# Help with a seemingly simple math problem!

• Aug 12th 2007, 11:38 PM
konnie
Help 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.

• Aug 13th 2007, 12:29 AM
CaptainBlack
Quote:

Originally Posted by konnie
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.

First do a diagram, see attachment:

RonL
• Aug 13th 2007, 02:57 AM
ticbol
Quote:

Originally Posted by konnie
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.

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

-----------------
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, 06:58 AM
Soroban
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 $y=3$ and has an x-intercept at $x=1$.

(a) Determine the other x-intercept
(b) Deduce the equation of the circle.

The center of the circle is: . $C(h,3)$
The y-intercept is: . $A(0,3)$
The x-intercept is: . $B(1,0)$

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

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

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

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

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

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

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

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

• Aug 15th 2007, 12:01 AM
konnie
Thank you ^___^