Find the equation to the circle which has its centre at the point (3,4) and touch the straight line 5x 12y=1
You are missing an operator in your line equation, I will assume you mean
5x + 12y = 1 or 5x + 12y - 1 = 0
we know that the equation must be
(x-3)^2 + (y-4)^2 = r^2 , so the only thing missing is the value of r
using the distance from a point to a line method,
r = |5(3) + 12(4)-1 |/√(3^2 + 4^2) = 62/5
(x-3)^2 + (y-4)^2 = 3844/25
To find the equation of a circle, we need the center of the circle and its radius. In this case, the center of the circle is given as (3, 4).
First, let's find the radius of the circle. The radius can be determined by finding the distance between the center of the circle and any point on the circle. Since the line 5x + 12y = 1 is tangent to the circle, the distance between the center of the circle and the line will be equal to the radius of the circle.
The general form of a line is Ax + By + C = 0. So, let's convert the given line to this form.
5x + 12y = 1
Rearranging the terms, we get:
12y = -5x + 1
y = (-5/12)x + 1/12
The line is now in the form y = mx + c, where m is the slope and c is the y-intercept.
The distance between a point (x1, y1) and a line Ax + By + C = 0 can be calculated using the formula:
distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2)
For our case, using the formula, the distance between the center (3, 4) and the line 5x + 12y = 1 is:
distance = |(5*3) + (12*4) - 1| / sqrt(5^2 + 12^2)
= |15 + 48 - 1| / sqrt(25 + 144)
= 62 / sqrt(169)
= 62 / 13
= 4.77 (approx.)
Now that we have the radius of the circle (4.77), we can write the equation for the circle. The general form of a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
Substituting the values, we get:
(x - 3)^2 + (y - 4)^2 = (4.77)^2
Simplifying further, we have:
(x - 3)^2 + (y - 4)^2 = 22.7529
Therefore, the equation of the circle with its center at (3, 4) and tangent to the line 5x + 12y = 1 is:
(x - 3)^2 + (y - 4)^2 = 22.7529