Answer:
The standard form of the given circle is
[tex](x-2)^2+(y-\frac{7}{2})^2=73[/tex]
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 [tex]M=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})[/tex]
Let [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] be the given points (6,2) and (-2,5) respectively.
[tex]M=(\frac{6-2}{2},\frac{2+5}{2})[/tex]
[tex]M=(\frac{4}{2},\frac{7}{2})[/tex]
[tex]M=(2,\frac{7}{2})[/tex]
Therefore the center (h,k) is [tex](2,\frac{7}{2})[/tex]
now to find the radius:
The diameter is the distance between the given points (6,2) and (-2,5)
[tex]d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
[tex]=\sqrt{(-2-6)^2+(5-2)^2}[/tex]
[tex]=\sqrt{(-8)^2+(3)^2}[/tex]
[tex]=\sqrt{64+9}[/tex]
[tex]=\sqrt{73}[/tex]
Therefore the radius is [tex]\sqrt{73}[/tex]
i.e., [tex]r=\sqrt{73}[/tex]
Therefore the standard form of the circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Now substituting the center and radiuswe get
[tex](x-2)^2+(y-\frac{7}{2})^2=(\sqrt{73})^2[/tex]
[tex](x-2)^2+(y-\frac{7}{2})^2=73[/tex]
Therefore the standard form of the given circle is
[tex](x-2)^2+(y-\frac{7}{2})^2=73[/tex]
