Write the equation of a circle that is tangent to the x-axis at (4,0) and has the y-intercepts of -2 and -8.
2007-08-28 07:57:01 · 5 answers · asked by Rachael A 2 in Science & Mathematics ➔ Mathematics
The circle is tangent to the x-axis at (4,0) => its center is at (4,-r), where r is the radius, as the y-intersects are negative.
The equation in question is of the form:
(x-4)^2 + (y+r)^2 = r^2.
The circle has the y-intercepts of -2 and -8 => (0,-2) and
(0,-8) lie on the circle =>
16 + (-2+r)^2 = r^2 <=>
16 + 4 -4r + r^2 = r^2 <=>
20 = 4r <=>
r = 5.
The equation is:
(x-4)^2 + (y+5)^2 = 25
Obviously (0,-8) also satisfies this equation.
2007-08-28 08:18:45 · answer #1 · answered by Anonymous · 0⤊ 0⤋
If the tangent is at (4,0) the the radius is perpendicular to the x-axis. so one component of the circle centre is 4.
As the circle intercepts y at -2 and -8 then the mid-point is -5, follow this through for another radii to -5, the other circle centre component is -5.
So the circle centre is the point (4,-5)
As the circle has a tangent with the x-axis and the ceentre is at -5 (y) then the radius is '5'.
NB It's easiest if you draw out the diagram, then it can be seen how all the points plot out.
Eq'n
(x - 4)^2 + (y --5)^2 = 5^2
x^2 - 8x + 16 + y^2 +10y + 25 = 25
x^2 + y^2 - 8x +10y +16 = 0
2007-08-28 08:15:37 · answer #2 · answered by lenpol7 7 · 0⤊ 0⤋
(x-4)^2+(y+5)^2 = 25
Since the circle is tangent to the x-axis at (4,0) then the x -coordinate of the center of the circle must be 4. Since the y intercepts are -2and -8, the y-coordinate of the circle's center must be (-2-8)/2 = -5.
Now you can easily determine r as the distanc fro the center to the point (4,0) which is 5.
2007-08-28 08:20:28 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
Drawing a picture always helps. You'll be able to see where the center is, and then it's a simple matter to find the radius.
The center must be 4 units to the right of the y-axis, because that's where the circle's tangent to the x-axis.
The center must be 6 units below the x-axis, because that's halfway between the y-intercepts.
So, the center is at ( 4, -6 ). Since you know the circle touches the point ( 4 , 0 ), and that's 6 units from the center, the radius must be 6.
(x-h)^2 + (y-k)^2 = r^2
(x-4)^2 + (y+6)^2 = 36
2007-08-28 08:06:00 · answer #4 · answered by Mehoo 3 · 0⤊ 1⤋
center C(4,-5) and radius 5
(x-4)^2+(y+5)^2 =25
2007-08-28 08:07:22 · answer #5 · answered by santmann2002 7 · 1⤊ 0⤋