The equations of two circles are: $c_1 : x^2 + y^2 - 6x - 10y + 29 = 0$ $c_2 : x^2 + y^2 - 2x - 2y - 43 = 0$ (a) Write down the centre and radius-length of each circle - Leaving Cert Mathematics - Question 2 - 2012

To prove the circles are touching, we need to find the distance between their centers and compare it to the difference of their radii.

1. Calculate the distance between centers:

   $d = ext{sqrt}((3 - 1)^2 + (5 - 1)^2) = ext{sqrt}(4 + 16) = ext{sqrt}(20) = 2 ext{sqrt}(5)$

2. The difference of the radii is:

   $|r_1 - r_2| = | ext{sqrt}(5) - ext{sqrt}(45)| = | ext{sqrt}(5) - 3 ext{sqrt}(5)| = | -2 ext{sqrt}(5)| = 2 ext{sqrt}(5)$

Since the distance between the centers equals the difference of the radii, the circles touch internally.

To verify that (4, 7) is a common point, we substitute the coordinates into both circle equations.

1. For $c_1$:

   $4^2 + 7^2 - 6(4) - 10(7) + 29 = 0$

   $$$$

ightarrow 0 = 0$$

2. For $c_2$:

   $4^2 + 7^2 - 2(4) - 2(7) - 43 = 0$

   $$$$

ightarrow 0 = 0$$

Both equations hold true, confirming that (4, 7) is a common point.

To find the equation of the common tangent, we can use point-slope form from the given points:

1. Calculate the slope from $(3, 5)$ to $(4, 7)$:

   $ext{slope} = \frac{7 - 5}{4 - 3} = \frac{2}{1} = 2$

2. The slope of the tangent line is the negative reciprocal:

   $m = -\frac{1}{2}$

3. Using point-slope form and point $(4, 7)$:

   $y - 7 = -\frac{1}{2}(x - 4)$

   Rearranging gives:

   $y = -\frac{1}{2}x + 2 + 7$

   $y = -\frac{1}{2}x + 9$

Thus, the equation of the common tangent is $y = -\frac{1}{2}x + 9$.