Answer:
![x^{2} +y^{2}-6y-12x+43=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2By%5E%7B2%7D-6y-12x%2B43%3D0)
Step-by-step explanation:
The general form of the equation of a circle is given as:
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.
are the two points.
Midpoint of (5, 4) and (7, 2) = ![\frac{1}{2}(5+7 , 4+2) =\frac{1}{2}(12 , 6)=(6,3)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%285%2B7%20%2C%204%2B2%29%20%3D%5Cfrac%7B1%7D%7B2%7D%2812%20%2C%206%29%3D%286%2C3%29)
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=![\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \\](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_1-x_2%29%5E2%2B%28y_1-y_2%29%5E2%7D%20%5C%5C)
=![\sqrt{(5-6)^2+(4-3)^2} \\=\sqrt{(-1)^2+(1)^2}=\sqrt{1+1}=\sqrt{2}\\r=\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B%285-6%29%5E2%2B%284-3%29%5E2%7D%20%5C%5C%3D%5Csqrt%7B%28-1%29%5E2%2B%281%29%5E2%7D%3D%5Csqrt%7B1%2B1%7D%3D%5Csqrt%7B2%7D%5C%5Cr%3D%5Csqrt%7B2%7D)
The equation of the circle centre (6,3) with radius
is therefore given as:
![(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](https://tex.z-dn.net/?f=%28x-a%29%5E2%2B%28y-b%29%5E2%3Dr%5E2%5C%5C%28x-6%29%5E2%2B%28y-3%29%5E2%3D%28%5Csqrt%7B2%7D%29%20%5E2%5C%5Cx%5E%7B2%7D%20%2By%5E%7B2%7D-6y-12x%2B45%3D2%5C%5Cx%5E%7B2%7D%20%2By%5E%7B2%7D-6y-12x%2B45-2%3D0%5C%5Cx%5E%7B2%7D%20%2By%5E%7B2%7D-6y-12x%2B43%3D0)