Find the equation of the tangent to the following circles at the given coordinates:
1) x^2 +y^2 = 25; (3;4)
2) (x - 1)^2 + (y - 2)^2 = 16; (1;6)
3) x^2 + 4x + y^2 - 6y = -4; (-2;6)
4) (x - 2)^2 + (y + 5)^2 = -20; (a;b)
1) To find the equation of the tangent to the circle x^2 +y^2 = 25 at the point (3,4), we first need to find the slope of the tangent line.
The slope of the tangent to a circle at a given point is equal to the negative reciprocal of the slope of the radius passing through that point.
Since the center of the circle is at the origin (0,0), the radius passing through the point (3,4) is a line segment from the origin to the point (3,4).
The slope of this radius can be found using the formula (y2 - y1) / (x2 - x1):
m = (4 - 0) / (3 - 0) = 4/3
The negative reciprocal of 4/3 is -3/4.
So the slope of the tangent line is -3/4.
Now we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
y - 4 = (-3/4)(x - 3)
y - 4 = (-3/4)x + 9/4
y = (-3/4)x + 25/4
Therefore, the equation of the tangent to the circle x^2 +y^2 = 25 at the point (3,4) is y = (-3/4)x + 25/4.
2) Similarly, to find the equation of the tangent to the circle (x - 1)^2 + (y - 2)^2 = 16 at the point (1,6), we need to find the slope of the tangent line.
The center of the circle is at the point (1,2), and the radius passing through the point (1,6) is a line segment from the center to the point.
The slope of this radius can be found using the formula (y2 - y1) / (x2 - x1):
m = (6 - 2) / (1 - 1) = 4/0, which is undefined.
Since the slope is undefined, the tangent line is a vertical line. The equation of a vertical line passing through the point (1,6) is x = 1.
Therefore, the equation of the tangent to the circle (x - 1)^2 + (y - 2)^2 = 16 at the point (1,6) is x = 1.
3) To find the equation of the tangent to the circle x^2 + 4x + y^2 - 6y = -4 at the point (-2,6), we need to find the slope of the tangent line.
First, we need to rearrange the equation of the circle to find the center and radius. Completing the square for the x and y terms, we get:
(x^2 + 4x) + (y^2 - 6y) = -4
(x^2 + 4x + 4) + (y^2 - 6y + 9) = -4 + 4 + 9
(x + 2)^2 + (y - 3)^2 = 9
So the center of the circle is at (-2,3) and the radius is 3.
The radius passing through the point (-2,6) is a line segment from the center to the point.
The slope of this radius can be found using the formula (y2 - y1) / (x2 - x1):
m = (6 - 3) / (-2 - (-2)) = 3/0, which is undefined.
Since the slope is undefined, the tangent line is a vertical line. The equation of a vertical line passing through the point (-2,6) is x = -2.
Therefore, the equation of the tangent to the circle x^2 + 4x + y^2 - 6y = -4 at the point (-2,6) is x = -2.
4) The equation (x - 2)^2 + (y + 5)^2 = -20 does not represent a valid circle because the sum of two positive squares cannot be negative. Therefore, there is no tangent line for this equation.