Find the equation of a circle with center C(1,6) and tangent line x-y-1=0.
equation of circle:
(x-1)^2 + (y-6)^2 = r^2
tangent line is y = x-1
When intersected , it should give us one root
(x-1)^2+ (x-1 - 6)^2 = r^2
x^2 - 2x + 1 + x^2 -14x + 47 - r^2 = 0
2x^2 - 16x + 48-r^2
a quadratic where
a = 2
b = -16
c = 48-r^2
to have only one root, b^2 - 4ac=0
256 - 4(2)(48-r^2) = 0
256 - 384 + 8r^2 = 0
r^2 = 16
(x-1)^2 + (y-6)^2 = 16
easier way:
distance from (1,6) to x-y-1 = 0 is
| (1-6-1) |/√(1+1) = 6/√2
r = 6/√2
r^2 = 36/2 = 16
equation: (x-1)^2 + (y-6)^2 = 16
To find the equation of a circle with a given center and a tangent line, we need to determine two things: the radius of the circle and the coordinates of a point on the circle.
Step 1: Calculate the distance from the center of the circle, C(1,6), to the tangent line x-y-1=0.
The distance from a point to a line can be found by the formula:
distance = |Ax + By + C| / √(A^2 + B^2)
Given the equation of the line x - y - 1 = 0, we can rewrite it in the standard form Ax + By + C = 0, where A = 1, B = -1, and C = -1.
Plugging in the values, we get:
distance = |1(1) + (-1)(6) + (-1)| / √(1^2 + (-1)^2)
= |1 - 6 - 1| / √(1 + 1)
= |-6| / √2
= 6 / √2
= 6√2 / 2
= 3√2
So, the radius of the circle is 3√2.
Step 2: Determine the coordinates of a point on the circle by finding the intersection of the line x-y-1=0 and the perpendicular line passing through C(1,6).
Since the given line has a slope of 1, the perpendicular line will have a slope of -1/1 = -1 (negative reciprocal).
The equation of the perpendicular line passing through point C(1,6) can be found using the point-slope form: y - y1 = m(x - x1), where (x1, y1) = (1,6) and m = -1.
y - 6 = -1(x - 1)
y - 6 = -x + 1
y = -x + 7
Now, we need to find the intersection point by solving the system of equations:
x - y - 1 = 0 (given tangent line)
y = -x + 7 (perpendicular line)
Substituting the value of y from the second equation into the first equation:
x - (-x + 7) - 1 = 0
x + x - 7 - 1 = 0
2x - 8 = 0
2x = 8
x = 4
Substituting the value of x into the second equation to find y:
y = -4 + 7
y = 3
Therefore, the intersection point is P(4,3).
Step 3: Write the equation of the circle using the center and radius.
The equation of a circle with center C(h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
Plugging in the values, we get:
(x - 1)^2 + (y - 6)^2 = (3√2)^2
(x - 1)^2 + (y - 6)^2 = 18
Thus, the equation of the circle is (x - 1)^2 + (y - 6)^2 = 18.
To find the equation of a circle with a given center and a tangent line, we need to understand a few concepts.
1. The center-radius form equation of a circle is given by:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) represents the center of the circle, and r represents the radius of the circle.
2. The tangent line to a circle is perpendicular to the radius at the point of tangency. Therefore, we can find the slope of the radius by taking the negative reciprocal of the slope of the tangent line.
Now, let's find the equation of the circle with the given center C(1,6) and the tangent line x - y - 1 = 0.
Step 1: Find the slope of the tangent line.
The given equation is x - y - 1 = 0.
To rewrite it in slope-intercept form (y = mx + c), we solve for y:
y = x - 1.
Comparing it with y = mx + c, we can see that the slope of the tangent line is 1.
Step 2: Find the slope of the radius.
The radius is perpendicular to the tangent line, so the slope of the radius is the negative reciprocal of the slope of the tangent line.
The negative reciprocal of 1 is -1.
Step 3: Find the equation of the circle using the center and radius.
Given the center C(1,6) and the slope of the radius -1, we can write the equation of the circle as:
(x - 1)^2 + (y - 6)^2 = r^2.
To find the value of r^2, we need to use the fact that the radius is the distance between the center and any point on the circle.
Step 4: Find the distance between the center and the tangent line.
The distance between the center C(1,6) and the tangent line x - y - 1 = 0 can be found using the formula for the distance between a point and a line:
d = |Ax + By + C| / sqrt(A^2 + B^2)
In this case, the equation of the tangent line can be rewritten as -x + y + 1 = 0, which represents the line Ax + By + C = 0 with A = -1, B = 1, and C = 1.
Plugging in these values, we have:
d = |-1*1 + 1*6 + 1| / sqrt((-1)^2 + 1^2) = 6 / sqrt(2)
Since the radius is perpendicular to the tangent line, the distance between the center and the tangent line is equal to the radius.
Step 5: Substitute the value of r into the equation of the circle.
We found in Step 4 that r = 6 / sqrt(2). Substituting this into the equation from Step 3, we have:
(x - 1)^2 + (y - 6)^2 = (6 / sqrt(2))^2
Simplifying further, we get:
(x - 1)^2 + (y - 6)^2 = 18
Therefore, the equation of the circle with center C(1,6) and tangent line x - y - 1 = 0 is (x - 1)^2 + (y - 6)^2 = 18.