If the end points of the diameter of a circle are (8,6) and (2,0), what is the standard form equation of the circle

Accepted Solution

Answer:[tex]\large\boxed{(x-5)^2+(y-3)^2=5^2\to(x-5)^2+(y-3)^2=25}[/tex]Step-by-step explanation:The standard form of an equation of a circle:[tex](x-h)^2+(y-k)^2=r^2[/tex](h, k) - centerr - radiusWe have the end points of a diameter (8, 6) and (2, 0). The midpoint of a diameter is a center of a circle. Half of a length of diameter is a length of a radius.The formula of a midpoint of a segment:[tex]\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)[/tex]Substitute:[tex]x=\dfrac{8+2}{2}=\dfrac{10}{2}=5\\\\y=\dfrac{6+0}{2}=\dfrac{6}{2}=3[/tex]The center of a circle is (5, 3).The formula of a length of a segment:[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]Substitute:[tex]d=\sqrt{(2-8)^2+(0-6)^2}=\sqrt{(-8)^2-(-6)^2}=\sqrt{64+36}=\sqrt{100}=10[/tex]The length of a diameter is 10 units. The length of a radius r = 10 : 2 = 5 units.Finally we have the equation of a circle:[tex](x-5)^2+(y-3)^2=5^2[/tex]