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!
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 $\displaystyle y=3$ 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: .$\displaystyle \boxed{{\color{blue}(9,\,0)}}$