Consider the circles x^2+y^2=1 and x^2+y^2-2x-6y+6=0. Then find the equation of a common tangents to a circles ???
the line joining the two centers is y=3x
the two tangents cross midway between where the line intersects the circles.
Now find where a line with slope -1/y' goes through that point.
Or, google the problem and I'm sure you will find a more general method.
My answer is 4x-3y-5=0
Ignore my previous post. But I don't think your equation is right either.
Consider that the two circles have
center (0,0) radius=1
center (1,3) radius=2
Then one common tangent is the line y=1
So the other will have negative slope, not a slope of 4/3
wow - my bad. I was thinking only of the internal tangents.
Good job.
To find the equation of the common tangents to the two circles, follow these steps:
Step 1: Simplify the equation of each circle to determine their center and radius:
For the circle x^2 + y^2 = 1:
Center: (0, 0)
Radius: √1 = 1
For the circle x^2 + y^2 - 2x - 6y + 6 = 0:
Complete the square to simplify the equation:
(x^2 - 2x) + (y^2 - 6y) = -6
(x^2 - 2x + 1) + (y^2 - 6y + 9) = -6 + 1 + 9
(x - 1)^2 + (y - 3)^2 = 4
Center: (1, 3)
Radius: √4 = 2
Step 2: Find the distance between the two centers (0, 0) and (1, 3):
Distance = √((1 - 0)^2 + (3 - 0)^2) = √(1 + 9) = √10
Step 3: Determine the relative position of the two circles and the number of common tangents:
Since the distance between the centers, √10, is greater than the sum of the radii (1 + 2 = 3), the circles do not intersect. Therefore, there are 4 common tangents that are external to the two circles.
Step 4: Find the equation for the common tangents:
For external common tangents, the formula is:
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
We'll use the equation of the circle x^2 + y^2 = 1 as the first circle.
(x - 0)(x - 1) + (y - 0)(y - 3) = 1
x(x - 1) + y(y - 3) = 1
Thus, the equation of the common tangents to the circles x^2 + y^2 = 1 and x^2 + y^2 - 2x - 6y + 6 = 0 is:
x(x - 1) + y(y - 3) = 1.