A circle has centre(5,12) and its tangent to the line with equation 2x-y+3=0. Wtite equation of the cirle
The distance from (5,12) to the line is
|2*5 - 1*12 + 3|/√(2^2+1^2) = 1/√5
So, the circle is
(x-5)^2 + (y-12)^2 = 1/5
check: The line and the circle must intersect at a single point.
(x-5)^2 + ((2x+3)-12)^2 = 1/5
This has a solution only at x = 23/5
To find the equation of the circle, we need to know its radius. The radius can be determined by finding the perpendicular distance between the center of the circle and the given line.
The given line has the equation 2x - y + 3 = 0.
The equation of a line in slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.
To put the given line equation in slope-intercept form, we solve for y:
2x - y + 3 = 0
-y = -2x - 3
y = 2x + 3
From the slope-intercept form, we can see that the slope of the given line is 2.
Since the line is tangent to the circle, the radius of the circle will be the perpendicular distance from the center (5,12) to the line 2x - y + 3 = 0.
The perpendicular distance between a point (x1, y1) and a line Ax + By + C = 0 is given by the formula:
d = |Ax1 + By1 + C| / sqrt(A^2 + B^2)
In our case, the center of the circle is (5,12), so x1 = 5 and y1 = 12. The equation of the line is 2x - y + 3 = 0, so A = 2, B = -1, and C = 3.
Calculating the distance:
d = |2(5) + (-1)(12) + 3| / sqrt(2^2 + (-1)^2)
= |10 - 12 + 3| / sqrt(4 + 1)
= |-1| / sqrt(5)
= 1 / sqrt(5)
Thus, the radius of the circle is 1 / sqrt(5).
Now we can write the equation of the circle using the center (5,12) and the radius:
(x - 5)^2 + (y - 12)^2 = (1 / sqrt(5))^2
Simplifying the equation:
(x - 5)^2 + (y - 12)^2 = 1 / 5
This is the equation of the circle.
To find the equation of the circle, we need to know its center and radius.
Given that the circle is tangent to the line with equation 2x - y + 3 = 0, we can find the perpendicular distance from the center of the circle to the line. This distance will be equal to the radius of the circle.
Step 1: Find the slope of the line.
Rewriting the equation of the line in slope-intercept form (y = mx + c), we have:
2x - y + 3 = 0
-y = -2x - 3
y = 2x + 3
Comparing this with the equation of a line (y = mx + c), we can see that the slope of the line is 2.
Step 2: Find the equation of a line perpendicular to the given line.
The slope of the line perpendicular to the given line is the negative reciprocal of its slope. Therefore, the slope of the perpendicular line is -1/2.
Step 3: Find the equation of the perpendicular line passing through the center of the circle.
Since the circle is tangent to the line, the center of the circle lies on the perpendicular line. We can use the point-slope form to find the equation of the perpendicular line.
Using the point-slope form (y - y1 = m(x - x1)), where (x1, y1) is the center of the circle, we have:
(y - 12) = (-1/2)(x - 5)
2(y - 12) = -x + 5
2y - 24 = -x + 5
x + 2y = 29
Therefore, the equation of the perpendicular line passing through the center of the circle is x + 2y = 29.
Step 4: Find the perpendicular distance from the center of the circle to the line.
Using the formula for the perpendicular distance from a point (x1, y1) to the line Ax + By + C = 0, which is given by:
Distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2)
Plugging in the values, we have:
Distance = |(2)(5) + (-1)(12) + 3| / sqrt(2^2 + (-1)^2)
Distance = |10 - 12 + 3| / sqrt(4 + 1)
Distance = |-2 + 3| / sqrt(5)
Distance = 1 / sqrt(5)
Step 5: The radius of the circle is equal to the perpendicular distance obtained in Step 4.
Therefore, the radius of the circle is 1 / sqrt(5).
Step 6: Write the equation of the circle.
The equation of a circle with center (h, k) and radius r is given by:
(x-h)^2 + (y-k)^2 = r^2
Plugging in the values, we have:
(x - 5)^2 + (y - 12)^2 = (1 / sqrt(5))^2
(x - 5)^2 + (y - 12)^2 = 1/5
Thus, the equation of the circle is (x - 5)^2 + (y - 12)^2 = 1/5.