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Find the standard form of the equation of each circle.

Circle with endpoints of a diameter at (4,3) and (0,1)

Short Answer

Expert verified

The standard form of the equation of a circle with endpoints of a diameter at (4,3) and (0, 1) is(x-2)2+(y-2)2=5


Step by step solution

01

Step 1. Given information

Here given that circle with endpoints of a diameter at (4,3) and (0,1).

We need to find the standard form of the equation of a circle.

02

Step 2. Explanation about finding  radius r of the circle

According to the question circle endpoint with a diameter is

(4,3) and (0,1) . The center of the circle is the midpoints of the circle endpoints (4,3) and (0,1) , hence the center is (4+02,3+12)=(2,2),weknowthatmidpoint(x,y)isbetween(x1,y1)and(x2,y2)andthen(x,y)=(x1+x22,y1+y22) The radius is the distance between the endpoint to the center, thus radius r is

r=(4-2)2+(3-2)2=(2)2+1=4+1=5

03

Step 3. Formation of standard of the equation of the circle

The standard form of the equation of a circle is of radius r with center at the

origin (h, k) is (x-h)2+(y-k)2=r2

Thus the equation of the given circle is(x-2)2+(y-2)2=52(x-2)2+(y-2)2=5

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