what is the equation of the circle whose centre lies on the line x-2y+2=0 and which touches the positive axes.
If it is tangent to the +x and +y axes, the center must also lie along the line
x = y, so that it is equidistant from both axes. The two lines cross at
y -2y +2 = 0, where y = 2.
Therefore x = y = 2 at the center. The radius is R = 2, for tangency.
The equation of the circle is
(x-2)^2 + (y-2)^2 = R^2 = 4
To find the equation of the circle, we need to determine the coordinates of its center and its radius.
1. Start by finding the coordinates of the center of the circle. Since the center lies on the line x - 2y + 2 = 0, we can rearrange this equation to solve for x in terms of y:
x = 2y - 2
2. To find the y-coordinate of the center, substitute x into the equation for the positive x-axis:
x = r, where r is the radius of the circle.
Since the circle touches the positive x-axis, the y-coordinate of the center is r.
3. Substitute the value of y into the equation (x = 2y - 2) to get the x-coordinate of the center:
x = 2(r) - 2
x = 2r - 2
4. The radius of the circle can be found by considering that it touches the positive axes. Since the circle only touches the positive x-axis, the radius is equal to r.
5. Now we have the coordinates of the center (x, y) = (2r - 2, r) and the radius r.
6. The general equation of a circle with center (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2
7. Substituting the center coordinates and radius into the equation, we have:
(x - (2r - 2))^2 + (y - r)^2 = r^2
Therefore, the equation of the circle whose center lies on the line x - 2y + 2 = 0 and which touches the positive axes is:
(x - (2r - 2))^2 + (y - r)^2 = r^2
To find the equation of the circle, we need to find the coordinates of its center and the radius.
The given line, x - 2y + 2 = 0 can be rewritten as x = 2y - 2.
Now, let's consider the possible centers of the circle. Since the circle touches the positive axes (x-axis and y-axis), the center will have coordinates (r, 0) or (0, r), where r is the radius of the circle.
Let's solve for the first possibility, center coordinates (r, 0):
Substituting x = r and y = 0 in the equation of the line:
r - 2(0) + 2 = 0
r + 2 = 0
r = -2
So, the center coordinates are (-2, 0).
Next, let's find the radius. Since the circle touches the positive x-axis, the distance from the center (-2, 0) to the x-axis should be equal to the radius. The distance from a point (x, y) to the x-axis is given by the formula |y|.
Distance from (-2, 0) to the x-axis = |0| = 0
Therefore, the radius of the circle is 0.
Now we have all the information needed to write the equation of the circle:
Center coordinates: (-2, 0)
Radius: 0
The equation of a circle with center coordinates (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2
Substituting the values from above:
(x - (-2))^2 + (y - 0)^2 = 0^2
(x + 2)^2 + y^2 = 0
Simplifying, we get:
(x + 2)^2 + y^2 = 0
Therefore, the equation of the circle whose center lies on the line x - 2y + 2 = 0 and which touches the positive axes is (x + 2)^2 + y^2 = 0.