A circle is tangent to the y-axis at y=3 and has one intercept at x=1.
a) Determine the other x-intercept.
b) Find the equation of the circle.
Thanks for helping!
To find the other x-intercept of the circle, we can use the fact that the circle is tangent to the y-axis at y=3. Since the y-intercept of the circle is at y=3, we know that the center of the circle lies on the line y=3.
Now, we are given that the circle has one intercept at x=1. This means that the circle intersects the x-axis at x=1. Since the y-coordinate of the x-intercept is 0, we can set y=0 in the equation of the circle.
Let's assume that the center of the circle is at the coordinates (h, 3), where h is the x-coordinate of the center. Then the equation of the circle can be written as:
(x - h)^2 + (y - 3)^2 = r^2
Since the circle passes through the point (1, 0), we can substitute these values into the equation:
(1 - h)^2 + (0 - 3)^2 = r^2
Simplifying this equation gives us:
1 - 2h + h^2 + 9 = r^2
Now, we also know that the circle is tangent to the y-axis at y=3. This means that the distance between the center of the circle and the y-axis is equal to the radius. In other words, the value of r is equal to h.
Substituting r=h into the equation, we have:
1 - 2h + h^2 + 9 = h^2
Simplifying this equation further:
-2h + 10 = 0
Solving for h gives us h = 5.
Now, we can substitute this value of h into the equation and solve for r:
1 - 2(5) + (5)^2 + 9 = r^2
1 - 10 + 25 + 9 = r^2
25 = r^2
Taking the square root of both sides gives us r = 5.
Therefore, the other x-intercept of the circle is x = -9 (using the symmetry of the circle) and the equation of the circle is (x-5)^2 + (y-3)^2 = 25.
This means the circle has its center at coordinates (5, 3) and a radius of 5.
a) To find the other x-intercept, we need to use the fact that the circle is tangent to the y-axis at y=3.
Since the circle is tangent to the y-axis, the distance from the center of the circle to the y-axis is equal to the radius of the circle. The y-coordinate of the center is given as 3, so the center of the circle is (0, 3).
We know that one x-intercept is at x=1. Let's call the other x-intercept x=a.
To find the distance between the center of the circle (0, 3) and the point of tangency on the y-axis (0, 3), we can use the distance formula:
Distance = sqrt((x₂ - x₁)² + (y₂ - y₁)²)
Plugging in the coordinates, we get:
Distance = sqrt((0 - a)² + (3 - 3)²)
Distance = sqrt(a² + 0)
Since the distance between the center and the point of tangency is equal to the radius, we have:
a² = radius²
The radius is the distance between the center and the x-intercept at x=1. Let's call this distance r.
r = sqrt((1 - 0)² + (0 - 3)²)
r = sqrt(1² + 3²)
r = sqrt(1 + 9)
r = sqrt(10)
So, the equation becomes:
a² = (sqrt(10))²
a² = 10
Taking the square root of both sides, we get:
a = ±sqrt(10)
Therefore, the other x-intercept is x=±sqrt(10).
b) To find the equation of the circle, we can use the general form of the equation of a circle:
(x - h)² + (y - k)² = r²
Where (h, k) represents the coordinates of the center of the circle and r represents the radius.
From part a), we know that the center of the circle is (0, 3) and the radius is sqrt(10).
Plugging these values into the equation, we have:
(x - 0)² + (y - 3)² = (sqrt(10))²
x² + (y - 3)² = 10
Therefore, the equation of the circle is x² + (y - 3)² = 10.
Well, if it is tangent to the y axis at (0,3), then the center must be somewhere on the horizontal line y = 3 because the radius is perpendicular to the tangent.
So if center is at (a,3)
(x-a)^2 + (y-3)^2 = r^2
we know when x = 0, y = 3 so
a^2 = r^2
so
(x-a)^2 + (y-3)^2 = a^2
now when y = 0, x = 1 and something else
(1-a)^2 +9 = a^2
1 -2a +a^2 +9 = a^2
a = 5 = r
so center is at (5,3)
(x-5)^2 + (y-3)^2 = 25
is the equation
x intercepts:
(x-5)^2 +9 = 25
x^2 - 10 x + 25 + 9 = 25
x^2 -10 x = -9
x^2 - 10 x + 25 = 16
(x-5)^2 = 16
x-5 = +/- 4
x = 1 or x = 9
which you could have seen from a sketch once you knew the radius was 5 and the center at (5,3)