The Problem
Our goal is to find the equations of the transverse common tangents of the two circles given by the following equations:
Circle 1: \(x^2 + y^2 + 4x - 6y + 4 = 0\)
Circle 2: \(x^2 + y^2 - 4x - 10y + 28 = 0\)
The visualization on the left shows these two circles plotted on a graph. Use the 'Next' button to walk through the solution step-by-step. Each step in the solution will correspond to a change in the visualization.
Step 1: Find Centers and Radii
First, we analyze the general equation of each circle, \(x^2 + y^2 + 2gx + 2fy + c = 0\), to find its center \((-g, -f)\) and radius \(r = \sqrt{g^2 + f^2 - c}\).
Circle 1
\(x^2 + y^2 + 4x - 6y + 4 = 0\)
Center \(O_1 = (-2, 3)\)
Radius \(r_1 = 3\)
Circle 2
\(x^2 + y^2 - 4x - 10y + 28 = 0\)
Center \(O_2 = (2, 5)\)
Radius \(r_2 = 1\)
On the chart, the centers of the circles, \(O_1\) and \(O_2\), are now marked.
Step 2: Find the Tangent Intersection Point (P)
For transverse tangents, the intersection point \(P\) divides the line segment connecting the centers \(O_1O_2\) internally in the ratio of the radii, \(r_1:r_2\), which is \(3:1\).
Using the internal division formula:
\(P(x,y) = \left(\frac{n x_1 + m x_2}{m+n}, \frac{n y_1 + m y_2}{m+n}\right)\)
We get \(P = \left(1, \frac{9}{2}\right)\).
The chart now shows the line segment connecting the centers and the crucial intersection point \(P\).
Step 3: General Equation of the Tangent
Any line passing through \(P(1, 9/2)\) can be written using the point-slope form \(y - y_p = m(x - x_p)\).
\(y - \frac{9}{2} = m(x - 1)\)
Rearranging this into the general form \(Ax + By + C = 0\), we get:
\(2mx - 2y + (9 - 2m) = 0\)
Our next task is to find the specific value(s) of the slope \(m\).
Step 4: Use the Distance Formula
The fundamental property of a tangent is that the distance from the center of a circle to the tangent line is equal to the circle's radius.
We'll set the distance from \(O_1(-2, 3)\) to our general tangent line equal to \(r_1 = 3\).
\(d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\)
Substituting our values leads to:
\(3 = \frac{|2m(-2) - 2(3) + (9 - 2m)|}{\sqrt{(2m)^2 + (-2)^2}}\)
After simplifying, we get the equation:
\(2\sqrt{m^2 + 1} = |1 - 2m|\)
Step 5: Solve for the Slope(s)
To solve for \(m\), we square both sides of the equation from the previous step:
\(4(m^2 + 1) = (1 - 2m)^2\)
\(4m^2 + 4 = 1 - 4m + 4m^2\)
This simplifies to \(4 = 1 - 4m\), which gives us \(m = -\frac{3}{4}\).
A Special Case: The Vertical Tangent
Our algebraic method only produced one slope. This often indicates the existence of a vertical tangent (with an undefined slope), which the \(y=mx+c\) form cannot represent. By checking if a vertical line \(x=1\) through point \(P\) is tangent to both circles, we find that it is!
Step 6: The Final Equations
We now have everything needed to state the equations of the two transverse common tangents.
Tangent 1 (with \(m = -3/4\))
\(3x + 4y - 21 = 0\)
Tangent 2 (the vertical line)
\(x = 1\)
The final tangent lines have been drawn on the chart, completing the solution visually.
