To find the equation of the circle that passes through the points [tex]\((3,2)\)[/tex] and [tex]\((5,4)\)[/tex] and has its center lying on the line [tex]\(3h - 2k - 1 = 0\)[/tex], we can follow these steps:
1. Equation of Circle and Key Variables:
Let the center of the circle be [tex]\((h, k)\)[/tex] and the radius be [tex]\(r\)[/tex].
The general equation of the circle is given by:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
2. Substitute Given Points:
Since the circle passes through the points [tex]\((3,2)\)[/tex] and [tex]\((5,4)\)[/tex], we can substitute these points into the circle's equation.
For point [tex]\((3,2)\)[/tex]:
[tex]\[
(3 - h)^2 + (2 - k)^2 = r^2 \quad \text{(Equation 1)}
\][/tex]
For point [tex]\((5,4)\)[/tex]:
[tex]\[
(5 - h)^2 + (4 - k)^2 = r^2 \quad \text{(Equation 2)}
\][/tex]
3. Line Condition:
Given that the center lies on the line [tex]\(3h - 2k - 1 = 0\)[/tex]:
[tex]\[
3h - 2k - 1 = 0 \quad \text{(Equation 3)}
\][/tex]
4. Solve the System of Equations:
We now have a system of three equations involving [tex]\(h\)[/tex], [tex]\(k\)[/tex], and [tex]\(r\)[/tex]:
[tex]\[
(3 - h)^2 + (2 - k)^2 = r^2
\][/tex]
[tex]\[
(5 - h)^2 + (4 - k)^2 = r^2
\][/tex]
[tex]\[
3h - 2k - 1 = 0
\][/tex]
Solving this system, we find:
[tex]\[
h = 3
\][/tex]
[tex]\[
k = 4
\][/tex]
[tex]\[
r = -2
\][/tex]
5. Equation of the Circle:
With [tex]\(h = 3\)[/tex], [tex]\(k = 4\)[/tex], and [tex]\(r = 2\)[/tex], the equation of the circle is:
[tex]\[
(x - 3)^2 + (y - 4)^2 = (-2)^2
\][/tex]
[tex]\[
(x - 3)^2 + (y - 4)^2 = 4
\][/tex]
Thus, the equation of the circle is:
[tex]\[
(x - 3)^2 + (y - 4)^2 = 4
\][/tex]
