Given
A circles passes through A(8,4) and B(0,-2).
To find the radius and the equation of the circle.
Explanation:
It is given that,
A circles passes through (8,4) and (0,-2).
Then, the diameter of the circle is,
[tex]\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ d=\sqrt{(8-0)^2+(4-(-2))^2} \\ =\sqrt{8^2+(4+2)^2} \\ =\sqrt{8^2+6^2} \\ =\sqrt{64+36} \\ =\sqrt{100} \\ =10\text{ }units \end{gathered}[/tex]Therefore, the radius of the circle is,
[tex]\begin{gathered} r=\frac{d}{2} \\ =\frac{10}{2} \\ =5\text{ }units \end{gathered}[/tex]Also, the center of the circle is,
[tex]\begin{gathered} Midpoint\text{ }of\text{ }AB=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ =(\frac{8+0}{2},\frac{4+(-2)}{2}) \\ =(\frac{8}{2},\frac{2}{2}) \\ =(4,1) \end{gathered}[/tex]Therefore, the center is (4,1).
Now, consider the equation of the circle as,
[tex](x-4)^2+(y-1)^2=5^2[/tex]That implies,
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