There are two tangents to the circle of radius centered at (0, - 1) that pass through the point (0,1). What are their equations?
Equation of a circle x^2+(y+1)^2=1^2
The radius is 1.
DRAW IT
Lets look at the right triangle on the right (remember tan hits circle at right angle to radius)
the lower leg is 1 the radius
the vertical leg (Hypotenuse) is 2 (from -1 to +1)
so the other leg is sqrt 3
The tangent to the circle is therefore sin^-1(1/2) or 30 degrees above the -y axis, 60 degrees below x axis
the slope m is therefore -sqrt 3/1
y = (-sqrt 3) x + b
when x = 0, y = 1so b= 1
y = (- sqrt 3)x + 1
on the left the slope is +sqrt 3
Clearly, the tangents have equation
y = 1±mx
for some slope m. So, taking m to be positive, the line meets the circle where
1-mx = √(1-x^2)-1
2-mx = √(1-x^2)
4-4mx+m^2x^2 = 1-x^2
(m^2+1)x^2 - 4mx + 3 = 0
To meet the circle in only one point, the discriminant must be zero, or
16m^2-12(m^2+1) = 0
4m^2 - 12 = 0
m = ±√3
so the lines are
y = 1±√3 x
see the graphs at
www.wolframalpha.com/input/?i=plot+x%5E2%2B%28y%2B1%29%5E2+%3D+1%2C+y+%3D+1-%E2%88%9A3+x%2C+y%3D1%2B%E2%88%9A3+x
To find the equations of the tangents, we will use the properties of tangents to a circle.
1. Find the slope of the line connecting the center of the circle and the point (0,1).
The center of the circle is (0, -1), so the slope of the line is (1 - -1)/(0 - 0) = 2/0, which is undefined.
2. Since the slope is undefined, the line connecting the center and the point (0,1) is a vertical line passing through (0, -1).
4. A tangent to a circle at a given point is perpendicular to the radius of the circle at that point.
5. The equation of the tangent line can be found using the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is the given point on the line and m is the slope.
6. Since the given point is (0,1) and the slope is undefined, the equation of the tangent line passing through (0,1) is x = 0.
7. Since there are two tangents to the circle, there is another tangent passing through (0,1) that is a vertical line parallel to the y-axis.
8. The equation of this tangent line is also x = 0.
Therefore, the equations of the two tangents to the circle of radius centered at (0, -1) that pass through the point (0,1) are x = 0.