Answer:
[tex]x^{2} +y^{2}-6y-12x+43=0[/tex]
Step-by-step explanation:
The general form of the equation of a circle is given as:
[tex](x-a)^2+(y-b)^2=r^2[/tex] where (a,b) are the centre and r is the radius.
If (5, 4) and (7, 2) are the endpoints of a diameter, we find the coordinate of the centre using the midpoint formula.
[tex]\frac{1}{2}(x_1+x_2 , y_1+y_2) \:where (x_1,y_1 \:and\: x_2, y_2)[/tex] are the two points.
Midpoint of (5, 4) and (7, 2) = [tex]\frac{1}{2}(5+7 , 4+2) =\frac{1}{2}(12 , 6)=(6,3)[/tex]
Next, we determine the radius. Using the distance formula, we find the distance from the centre to any of the endpoints.
Distance from (5,4) to (6,3)
r=[tex]\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \\[/tex]
=[tex]\sqrt{(5-6)^2+(4-3)^2} \\=\sqrt{(-1)^2+(1)^2}=\sqrt{1+1}=\sqrt{2}\\r=\sqrt{2}[/tex]
The equation of the circle centre (6,3) with radius [tex]\sqrt{2}[/tex] is therefore given as:
[tex](x-a)^2+(y-b)^2=r^2\\(x-6)^2+(y-3)^2=(\sqrt{2}) ^2\\x^{2} +y^{2}-6y-12x+45=2\\x^{2} +y^{2}-6y-12x+45-2=0\\x^{2} +y^{2}-6y-12x+43=0[/tex]
