Question

Write the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5).

Answer 1

Answer:

$(x+1)^2 + (y- 3.5)^2 =64$

Step-by-step explanation:

We are given that  endpoints of a diameter at the points (7,2) and (-9,5).

Center of the circle is the mid point of diameter

So, first find the mid point of diameter

Formula : $x=\frac{x_1+x_2}{2} , y=\frac{y_1+y_2}{2}$

$(x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)$

Substitute the values in formula

$x=\frac{7-9}{2} , y=\frac{2+5}{2}$

$x=-1 , y=3.5$

So, center of circle = (h,k )=(-1,3.5)

To find length of diameter :

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

$(x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)$

$d=\sqrt{(-9-7)^2+(5-2)^2}$

$d=\sqrt{265}$

Length of radius = r = $\frac{Diameter}{2}=\frac{\sqrt{256}}{2}$

Standard form of the equation of the circle : $(x -h)^2 + (y- k)^2 = r^2$

(h,k )=(-1,3.5)

$r=\frac{\sqrt{256}}{2}$

Equation of the circle : $(x+1)^2 + (y- 3.5)^2 =(\frac{\sqrt{256}}{2})^2$

Equation of the circle : $(x+1)^2 + (y- 3.5)^2 =64$

Hence  the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5) is  $(x+1)^2 + (y- 3.5)^2 =64$