Question

Find the standard form of the equation of the circle with endpoints of a diameter at the points (7,8) and (-3,6). Type the standard form of the equation of this circle​

Answer 1

Answer:

The equation of the circle is $(x - 2)^2 + (y - 7)^2 = 21$

Step-by-step explanation:

Equation of a circle:

The equation of a circle, with center $(x_0,y_0)$ and radius r is given by:

$(x - x_0)^2 + (y - y_0)^2 = r^2$

Distance between two points:

Suppose that we have two points, $(x_1,y_1)$ and $(x_2,y_2)$. The distance between them is given by:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Diameter at the points (7,8) and (-3,6).

The diameter is the distance between these two points, so:

$D = \sqrt{(-3-7)^2+(6-8)^2} = \sqrt{104}$

Radius is half the diameter, so:

$r = \frac{\sqrt{104}}{2} = \frac{\sqrt{104}}{\sqrt{4}} = \sqrt{\frac{104}{4}} = \sqrt{21}$

So

$r^2 = (\sqrt{21})^2 = 21$

Center:

Midpoint of the diameter, which is the mean of the coordinates. So

$x_0 = \frac{7 - 3}{2} = \frac{4}{2} = 2$

$y_0 = \frac{8 + 6}{2} = \frac{14}{2} = 7$

Then

$(x - 2)^2 + (y - 7)^2 = 21$