Find the equation to the circle which has its centre at the point (3,4) and touches the straight line 5x+12y=1
ahhh, see
http://www.jiskha.com/display.cgi?id=1437880873
Looks like I assumed correctly
To find the equation of the circle with its center at the point (3, 4) that touches the line 5x + 12y = 1, we can follow these steps:
1. Firstly, let's find the distance between the center of the circle and the line. The distance from a point (x0, y0) to a line Ax + By + C = 0 is given by the formula:
Distance = |Ax0 + By0 + C| / sqrt(A^2 + B^2)
In this case, the center of the circle is (3, 4), and the line is 5x + 12y = 1. Therefore, A = 5, B = 12, C = -1, x0 = 3, and y0 = 4. Substituting these values into the distance formula:
Distance = |5(3) + 12(4) - 1| / sqrt(5^2 + 12^2)
= |15 + 48 - 1| / sqrt(25 + 144)
= |62| / sqrt(169)
= 62 / 13
= 4.77 (rounded to two decimal places)
2. Since the circle touches the line, the distance between the center of the circle and the line is equal to the radius of the circle. Therefore, the radius of the circle is 4.77.
3. The equation of a circle with center (h, k) and radius r is given by the formula:
(x - h)^2 + (y - k)^2 = r^2
Substituting the values of the center (h, k) = (3, 4) and the radius r = 4.77 into the equation of the circle:
(x - 3)^2 + (y - 4)^2 = 4.77^2
Simplifying further:
(x - 3)^2 + (y - 4)^2 = 22.7929
Therefore, the equation of the circle is (x - 3)^2 + (y - 4)^2 = 22.7929.