find the equation of the circle having center (3,4) & touches the line 5x+12y+11=0
To find the equation of the circle that touches the line 5x+12y+11=0 and has a center at (3,4), we need to find the radius of the circle first.
The distance between the center of the circle and the given line is equal to the radius of the circle. We can use the formula for the distance from a point to a line to find the distance:
Distance = |Ax + By + C| / √(A^2 + B^2)
In this case, A = 5, B = 12, and C = 11. The center of the circle is (3,4), so x = 3 and y = 4. Plugging these values into the distance formula, we get:
Distance = |5(3) + 12(4) + 11| / √(5^2 + 12^2)
= |15 + 48 + 11| / √(25 + 144)
= 74 / √169
= 74 / 13
= 5.69 (approximately)
Therefore, the radius of the circle is approximately 5.69 units.
Now that we have the center (3,4) and the radius 5.69, we can write the equation of the circle in the general form:
(x - h)^2 + (y - k)^2 = r^2
Substituting the given values, we get:
(x - 3)^2 + (y - 4)^2 = (5.69)^2
Simplifying, we have:
(x - 3)^2 + (y - 4)^2 = 32.3761
So, the equation of the circle with center (3,4) and touches the line 5x+12y+11=0 is (x - 3)^2 + (y - 4)^2 = 32.3761.
the distance from the line to the point is |5*3+12*4+11|/√(5^2+12^2) = 74/13
So, now you know the center and the radius. Just write the equation.
see the graphs at
http://www.wolframalpha.com/input/?i=plot+5x%2B12y%2B11%3D0%2C+%28x-3%29^2%2B%28y-4%29^2+%3D+%2874%2F13%29^2