To find the transverse common tangents of the given circles, we'll follow a step-by-step approach to determine their centers, radii, and the conditions necessary for transverse tangents to exist.
Step 1: Rewrite each circle's equation in standard form.
For the circle [tex]\(x^2 + y^2 - 4x - 10y + 28 = 0\)[/tex]:
1.1. Group terms involving [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x^2 - 4x + y^2 - 10y = -28. \][/tex]
1.2. Complete the square for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ (x^2 - 4x) + (y^2 - 10y) = -28. \][/tex]
[tex]\[ (x^2 - 4x + 4) + (y^2 - 10y + 25) = -28 + 4 + 25. \][/tex]
[tex]\[ (x - 2)^2 + (y - 5)^2 = 1. \][/tex]
This is the standard form of a circle with center [tex]\((h1, k1) = (2, 5)\)[/tex] and radius [tex]\( r1 = \sqrt{1} = 1 \)[/tex].
For the circle [tex]\(x^2 + y^2 - 4x - 6y + 4 = 0\)[/tex]:
1.3. Group terms involving [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x^2 - 4x + y^2 - 6y = -4. \][/tex]
1.4. Complete the square for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ (x^2 - 4x) + (y^2 - 6y) = -4. \][/tex]
[tex]\[ (x^2 - 4x + 4) + (y^2 - 6y + 9) = -4 + 4 + 9. \][/tex]
[tex]\[ (x - 2)^2 + (y - 3)^2 = 9. \][/tex]
This is the standard form of a circle with center [tex]\((h2, k2) = (2, 3)\)[/tex] and radius [tex]\( r2 = \sqrt{9} = 3 \)[/tex].
Step 2: Calculate the distance between the centers of the circles.
[tex]\[ \text{Distance} \, d = \sqrt{(h2 - h1)^2 + (k2 - k1)^2} = \sqrt{(2 - 2)^2 + (3 - 5)^2} = \sqrt{0 + (-2)^2} = \sqrt{4} = 2. \][/tex]
Step 3: Check the condition for transverse common tangents.
Transverse common tangents exist if the sum of the radii is less than the distance between the centers:
[tex]\[ r1 + r2 < d. \][/tex]
Given:
[tex]\[ r1 = 1 \][/tex]
[tex]\[ r2 = 3 \][/tex]
[tex]\[ d = 2 \][/tex]
Checking the condition:
[tex]\[ r1 + r2 = 1 + 3 = 4. \][/tex]
[tex]\[ 4 \not< 2. \][/tex]
Since [tex]\(r1 + r2\)[/tex] is not less than [tex]\(d\)[/tex], there are no transverse common tangents for these circles.
Therefore, the final answer is:
No transverse common tangents exist.
