Answer:
The standard form of the given circle is
![(x-2)^2+(y-\frac{7}{2})^2=73](https://tex.z-dn.net/?f=%28x-2%29%5E2%2B%28y-%5Cfrac%7B7%7D%7B2%7D%29%5E2%3D73)
Step-by-step explanation:
Given that the end points of a diameter of a circle are (6,2) and (-2,5);
Now to find the standard form of the equation of this circle:
The center is (h,k) of the circle is the midpoint of the given diameter
midpoint formula is ![M=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})](https://tex.z-dn.net/?f=M%3D%28%5Cfrac%7Bx_%7B1%7D%2Bx_%7B2%7D%7D%7B2%7D%2C%5Cfrac%7By_%7B1%7D%2By_%7B2%7D%7D%7B2%7D%29)
Let
and
be the given points (6,2) and (-2,5) respectively.
![M=(\frac{6-2}{2},\frac{2+5}{2})](https://tex.z-dn.net/?f=M%3D%28%5Cfrac%7B6-2%7D%7B2%7D%2C%5Cfrac%7B2%2B5%7D%7B2%7D%29)
![M=(\frac{4}{2},\frac{7}{2})](https://tex.z-dn.net/?f=M%3D%28%5Cfrac%7B4%7D%7B2%7D%2C%5Cfrac%7B7%7D%7B2%7D%29)
![M=(2,\frac{7}{2})](https://tex.z-dn.net/?f=M%3D%282%2C%5Cfrac%7B7%7D%7B2%7D%29)
Therefore the center (h,k) is ![(2,\frac{7}{2})](https://tex.z-dn.net/?f=%282%2C%5Cfrac%7B7%7D%7B2%7D%29)
now to find the radius:
The diameter is the distance between the given points (6,2) and (-2,5)
![d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_%7B2%7D-x_%7B1%7D%29%5E2%2B%28y_%7B2%7D-y_%7B1%7D%29%5E2%7D)
![=\sqrt{(-2-6)^2+(5-2)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%28-2-6%29%5E2%2B%285-2%29%5E2%7D)
![=\sqrt{(-8)^2+(3)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%28-8%29%5E2%2B%283%29%5E2%7D)
![=\sqrt{64+9}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B64%2B9%7D)
![=\sqrt{73}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B73%7D)
Therefore the radius is ![\sqrt{73}](https://tex.z-dn.net/?f=%5Csqrt%7B73%7D)
i.e., ![r=\sqrt{73}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B73%7D)
Therefore the standard form of the circle is
![(x-h)^2+(y-k)^2=r^2](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2)
Now substituting the center and radiuswe get
![(x-2)^2+(y-\frac{7}{2})^2=(\sqrt{73})^2](https://tex.z-dn.net/?f=%28x-2%29%5E2%2B%28y-%5Cfrac%7B7%7D%7B2%7D%29%5E2%3D%28%5Csqrt%7B73%7D%29%5E2)
![(x-2)^2+(y-\frac{7}{2})^2=73](https://tex.z-dn.net/?f=%28x-2%29%5E2%2B%28y-%5Cfrac%7B7%7D%7B2%7D%29%5E2%3D73)
Therefore the standard form of the given circle is
![(x-2)^2+(y-\frac{7}{2})^2=73](https://tex.z-dn.net/?f=%28x-2%29%5E2%2B%28y-%5Cfrac%7B7%7D%7B2%7D%29%5E2%3D73)