Help explain the problem of this equation of circles .

Let the equation of the circle be: x 2 + y 2 + dx + ey + f = 0, then the coordinates of the intersection of the circle and the y axis are (0, x 2 + dx + f = 0) and the intersection coordinates of the x axis are (y 2 + ey + f = 0, 0) x 2 + dx + f = 0 The two intercepts of the circle on the x axis, and the sum of the two is -d

The two poles of y 2+ey+f=0 are the two intercepts of the circle on the y-axis, and the sum of the two is -e

So (-d-e)*2=4

So -d-e=2

The intercept of the circle on the x-axis is x 2 + dx + f = 0

The intercept of the circle on the y-axis is y2+ey+f=0

According to Weida's theorem x1 + x2 = - (coefficient of the primary term).

So, the sum of intercepts on the x-axis is -d

The intercepts on the y-axis and the sum are -e

So -d-e=4

Well, you're not mistaken.

A general equation for a circle. Then let y=0, we get x 2+dx+f=0, and the two roots of this equation are the intercepts of the circle on the x-axis, and the sum of the two is -dIn the same way, the sum of two roots on the y-axis is -e-d-e=2 is listed according to the title.

The general equation for a circle is.

x^2+y^2+dx+ey+f=0

Substitution points (4,2), (1,3).

16+4+4d+2e+f=0

1+9-d+3e+f=0

Excuse me for misunderstanding wrong.

If I understand correctly, it should be used.

x=0. y^2+ey+f=0

y1+y2=-e

y=0x^2+dx+f=0

x1+x2=-d

d+(-e)=4, good luck! Have fun!

As shown in the figure below, first write out the circular equation of the intersection point of the two circles, then find the coordinates of the center of the circle, and then substitute the coordinates of the center of the circle into the linear equation to get the coefficient, and then substitute the coefficient into the circular equation to get the circular equation, and sort it out.

The minimum value is 1, and the problem is very simple, three kinds of gong'qie'The xian male tangent line is the outer cut of two circles. Calculate the distance between the two circles to be 3, you can get a 2+4b 2=9

Question 1: x 2+y 2+2ax-4ay+5a 2-4=0 can be reduced to (x+a) 2+(y-2a) 2=4 From the meaning of the question a>0, the curve is a circle, the coordinates of the center of the circle are (-a, 2a), the distance from the center of the circle to the x-axis is 2a, the distance to the y-axis is a, the radius of the circle is 2, then a>2

Question 2: The coordinates of the center of the circle are (-1, -1), and the distance from the center of the circle to the straight line is d=|3(-1)+4(-1)-2)|/5 ， mn|The minimum value of is .

d-r=

The key to solving a system of equations is elimination.

For reference, please smile.

On x+y-1=0, the garden is tangent to x=y-1 at point a.

So the center of the circle is the intersection of the perpendicular line passing through point a and y=-2x.

Perpendicular equation: y=x-3

So the center o of the circle is (1,-2).

The equation for ao= 2 is: (x-1) +y+2) =2,3)*

The equation for a circle with a center of point c(a, b) and a radius of r: (x-a) 2 + y-b) 2 = r 2

1.The center of the circle is the point c(8,-3), (x-8) 2 +(y+3) 2 = r 2

and pass the point a(5,1), 5-8) 2 +(1+3) 2 = r 2 = 25, r = 5

Equation for a circle = (x-8) 2 +(y+3) 2 = 25

2.Pass a(-1,5), b(5,5), c(6,-2).

1-a)² 5-b)² = r² -1)

5-a)² 5-b)² = r² -2)

6-a)² 2-b)² = r² -3)

2)-(1) a -10a+25-a -2a-1=0, a = 2, substitution (1), 3).

4) 9 + 5-b)² = r² = b²-10b+25 +9 = b²-10b+34

5) 16 + 2-b)² = r² = b²+4b+4+16 = b²+4b+20

4)-(5) -14b + 14 = 0, b = 1, substitution (4).

9 + 16 = r², r = 5

x - 2)² y - 1)² = 25

Untie; 1，|ac|2=(8-5) 2+(-3-1) 2=25, so s=25 ;

2, let the equation for the circle be (x-a) 2+(y-b) 2=r 2 (-1-a) 2+(5-b) 2=r 2(5-a) 2+(5-b) 2=r 2

6-a)^2+(-2-b)^2=r^2

Solution: a=2, b=1, r 2=25

s=25π