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Alenkasestr [34]
3 years ago
11

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

Mathematics
1 answer:
lys-0071 [83]3 years ago
5 0

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

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