Knowing the two points P 1,3 and Q 2,1, we find the standard equation for a circle with the diameter

Let the center of the circle be o(a,b).

Then a=(-1+2) 2=1 2 , b=(3+1) 2=2 is centered on (1 2,2).

The length of the line segment pq is (1-2) 2+(3-1) 2= 13, so the radius is 13 2

Then the standard equation for a circle is (x-1 2) 2+(y-2) 2=13 4

There are many ways to solve this problem, and the simplest one is to use the vector method. Let any point on the circle a(x,y) then the vector p=(x+1,y-3), vector pb=(x-2,y-1), because pb is perpendicular, so the vector pa point multiplication vector pb=0, that is, (x+1)(x-2)+(y-3)(y-1)=0, simplified to obtain:

x 2-x+y 2-4y+1=0, which is the standard equation (x-1 2) 2+(y-2) 2=13 4

If the upper point of the garden is k(x,y), then the angle pkq=90 degrees.

kp^2+kq^2=pq^2

x+1)^2+(y-3)^2+(x-2)^2+(y-1)^2=(2+1)^2+(1-3)^2

x-1/2)^2+(y-2)^2=(3/2)^2

Center of the circle: x=(-1+2) 2=1 2, y=(3+1) 2=2

Radius: r = sqrt((2-1) 2 + 1-3) 2) 2=sqrt(5) 2

So the equation (x-1 2) 2 + y-2) 2 = 5 4

Summary. 3 The two points p(5,0) and q(1,2 5) are known, and the standard equation for the circle with the line segment pq as the diameter is obtained.

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p(5,0 ),q(1,25)

Center c = (3, 25 Zhen Bend Bu Miho 2).

r= |pq|/2 = 1/2)√(16+25) =1/2)√(41)

r^2 = 41/4

The line judgment pq is the standard equation for the diameter of the circle.

x-3)^2 +(y-25/2)^2 =41/4

|pq|=√1-2)²+3-(-1)]²5

The standard equation with the line segment pq as the radius and the point p as the center of the circle is (x+1) +y-3) =25

Let the center of the circle be o(a,b).

Then a=(-1+2) 2=1 2 , b=(3+1) 2=2 is centered on (1 2,2).

The length of the line segment pq is (1-2) 2+(3-1) 2= 13, so the radius is 13 2

Then the standard equation for a circle is (x-1 2) 2+(y-2) 2=13 4

Summary. Consultation Record · On 2022-12-132, + two points +p(3,1) and +q(5,0) are known, and the standard equation of a circle with the diameter of the line segment +pq+ is obtained.

2. + Knowing two points + p(3,1) and +q(5,0), find the standard equation of the circle with the line segment +pq+ as the diameter.

How to answer. That's how it was answered, dear.

Solution: and 2R 丨pq丨 (3-5) is absolutely stared at 2 ten (1-0) 2 5, the coordinates of the center of the circle are (4, 1 2), so the standard equation of the circle is:

xa4) 2 ten(xa1 2) 2 5 4.

The coordinates of the center o are: (3, 5).

aq=2√5，ap=5-1=4

pq²=aq²+ap²=20+16=36

pq=6，op=3

The standard Sun Zhun equation of the circle:

x-3) state envy base + (y- 5) = 9

Since pq is the diameter, the midpoint is the coordinate of the center of the circle, i.e., (5+1) 2=3, (0+2 5) 2= 5, and the center of the circle is (3, 5).

radius = pq 2= (1-5) +2 5) 2=3. Draft.

So the equation for the circle is (x-3) +y- 5) =9.

pq is the diameter, then the center of the circle is the midpoint of pq.

The center of the circle is (3, 5).

The square of the diameter is the sail thickness:

The square of the radius is:

The standard equation for a circle with the socks car line segment pq as the direct base diameter is:

（x-3）^2+(y-√5)^2=9