+23 Compute the length of the segment tangent from the origin to the circle that passes through the points (3, 4), (6, 8) and (5, 13).
+281 To find the length of the tangent from the origin to the circle, we first need to determine the equation of the circle. We can do this by using the three given points to form a system of equations.
Let the equation of the circle be:
(x-a)^2 + (y-b)^2 = r^2
Substituting the coordinates of the three points into this equation, we get a system of three equations:
(3-a)^2 + (4-b)^2 = r^2
(6-a)^2 + (8-b)^2 = r^2
(5-a)^2 + (13-b)^2 = r^2
Solving this system of equations, we find that the center of the circle is at (-2, -4/3) and the radius is 13√(28/9).
Now, to find the length of the tangent from the origin to the circle, we can use the Pythagorean theorem. Let's denote the length of the tangent as t, the distance from the origin to the center of the circle as d, and the radius of the circle as r. We have:
d^2 + t^2 = r^2
We can calculate d using the distance formula:
d = sqrt((-2)^2 + (-4/3)^2) = sqrt(68/9)
Substituting the values of d and r into the Pythagorean theorem equation, we get:
(68/9) + t^2 = (13√(28/9))^2
Solving for t, we find that the length of the tangent is equal to 10.
Therefore, the length of the segment tangent from the origin to the circle is 10.
