Question

I need help writing a equation for a circleEnds of diameter: (1,-2) and (10,8)

Answer 1

The equation of the circle with the given end point is;

`(x-5.5)^2+(y-3)^2=\text{ }(181)/(4)`

Here, we want to write the equation of a circle, given the end points of the circle

Given the end-points, we can use the formual for the distance to get the radius of the circle

We have this as follows;

`\begin{gathered} D\text{ = }\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \\ (x_1,y_1)\text{ = (1,-2)} \\ (x_2,y_2)\text{ = (10,8)} \\ D\text{ = }\sqrt[]{(10-1)^2+(8-(-2))^2} \\ D\text{ = }\sqrt[]{9^2+10^2} \\ D\text{ = }\sqrt[]{181} \end{gathered}`

To get the radius of the circle, we have to divide the diameter by 2

We have this as;

`r\text{ = }(D)/(2)\text{ = }\frac{\sqrt[]{181}}{2}`

Now, the other thing needed to write the equation of the circle is the center of the circle

The center of the circle can be calculated by the use of the midpoint formula

`\begin{gathered} (x,y)\text{ = (}(x_2+x_1)/(2),(y_2+y_1)/(2)) \\ \\ (x,y)\text{ = (}(1+10)/(2),(8-2)/(2))\text{ = (5.5,3)} \end{gathered}`

The general equation of the circle is;

`\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ (a,b)\text{ = (5.5,3)} \\ r\text{ = }\frac{\sqrt[]{181}}{2} \\ \\ (x-5.5)^2+(y-3)^2=\text{ }(\frac{\sqrt[]{181}}{2})^2 \\ \\ (x-5.5)^2+(y-3)^2=\text{ }(181)/(4) \end{gathered}`