The equation of the circle speed to find the answer all the formulas in the equation of the circle

The equation for the straight line ab is y=2x-3

The line segment ab perpendicularly bisects the linear line equation y=-x 2-1 2 center coordinates c(1 3,-2 3).

The radius of the circle c r=5 2 3

Find the equation for circle c (x-1 3) +y+2 3) =50 9, and let the distance from the center c of the circle c to the straight line l: x-y-m=0 be d

d=|m-1|/√2=√(50/9-1/2)m=(3±2√23)/3

a, b two-point straight line y=2x-3

a, b two points perpendicular bisector line y=-x 2-1 2 center of the circle (1 3, -2 3).

r*r=50/9

1) Find the circular c equation (x-1 3) 2+(y+2 3) 2=50 9

Let the radius r of the center of the circle (a, -2a).

Circle: (x-a) 2+(y+2a) 2=r 2ab substituting, get:

a=1/3 r^2=50/9

Circle: (x-1 3) 2+(y+2 3) 2=50 9 The distance from the center of the circle to the line l:x-y-m=0 is set to d, d=|1/3+2/3+m|/sqr(2)=|1+m|/genhao(2)

d^2+(|mn|2) 2=r 2 (Pythagorean theorem) Do the math yourself.

To be honest, no points and no motivation.

Is it still there! I'll do the math for you.

The ordinary equation of a circle：x²+y²+dx+ey+f=0; (d²+e²>4f)

The standard equation for a circle：（x-a)²+y-b)²=r²

Parametric equations for circles：x=a+rcosθ;y=b+rsin (as a parameter).

Tangent equation for a circle:

The tangent of a circle at the point above the circle x +y + dx + ey+f=0 (x0, y0) is x0x+y0y+ (x+x0)+ y+y0)+f=0

Formula for calculating circles:

1. The circumference of the circle c=2 r= d

2. The area of the circle s = r 2

3. The arc length of the sector l=n r 180

4. Sector area s=n r 2; /360=rl/2

5. The area of the conical side is s= rl

The center of the circle is at the origin, and the radius is r: x +y = r;

The center of the circle is in (a,b), and the radius is r: (x-a) +y-b) = r ;

The general equation of the garden: x + y + cx + dy+ m = 0 Features: The coefficients of the quadratic terms are equal; There are no crossovers; ③。m-c²/4-d²/4<0.

The parametric equation of the garden with the center of the circle in (a, b) and the radius r is: x=a+rcost; y=b+rsint；t∈r.

1. A circle with the center of the circle at the origin (0,0) and the radius of r is r.

x²+y²=r²

2. The center of the circle is in (a, b), and the radius is r

x-a)²+y-b)²=r²

3. General equation of the circle:

x²+y²+mx+ny+p=0

Features: a. The coefficients of the quadratic terms are equal.

b. There are no cross items.

c、p-m²/4-n²/4<0

4. The parametric equation of the circle with the center of the circle in (a, b) and the radius r is

x=a+rcost；y=b+rsint（t∈r）

High-quality answers.

The ordinary equation for a circle: x +y + dx + ey+f = 0; (d +e >4f) the standard equation for a circle: (x-a) +y-b) = r parametric equation for a circle:

x=a+rcosθ;The tangent equation for the circle of y=b+rsin (is parameter):

The tangent of a circle over a circle x +y + dx + ey+f=0 is x0x+y0y+ (x+x0)+ y+y0)+f=0 x +y =r The tangent equation for a circle at the previous point (x0,y0): x0x+y0y=r

1. A circle with the center of the circle at the origin (0,0) and the radius of r is r.

x²+y²=r²

2. The center of the circle is in (a, b), and the radius is r

x-a)²+y-b)²=r²

3. General equation of the circle:

x²+y²+mx+ny+p=0

Features: a. The coefficients of the quadratic terms are equal.

b. There are no cross items.

c、p-m²/4-n²/4<0

4. The parametric equation of the circle with the center of the circle in (a, b) and the radius r is

x=a+rcost；y=b+rsint（t∈r）

The ordinary equation for a circle: x +y + dx + ey+f = 0;

d²+e²>4f)

The standard equation for a circle: (x-a) +y-b) =r The parametric equation for a circle: x=a+rcos ;

y=b+rsinθ

is the tangent equation of the circle

The tangent of a circle at the point above the circle x +y + dx + ey+f=0 (x0, y0) is x0x+y0y+ (x+x0)+ y+y0)+f=0

Extended Materials. Formula for calculating circles:

1. The circumference of the circle c=2 r= d

2. The area of the circle s = r 2

3. The arc length of the sector l=n r 180

4. Sector area s=n r 2; 360 = RL 25, conical side area s = RL

The center of the circle is at the origin, and the radius is r: x +y = r;

The center of the circle is in (a,b), and the radius is r: (x-a) +y-b) = r ;

The general equation of the garden: x + y + cx + dy+ m = 0 Features: The coefficients of the quadratic terms are equal; There are no crossovers; ③。m-c²/4-d²/4<0.

The parametric equation of the garden with the center of the circle in (a, b) and the radius r is: x=a+rcost; y=b+rsint；t∈r.

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 center of the circle to the book of scattered stupidity.

The distance of the y-axis is equal to.

2, so x=2 or punch-2

Let y=kx+c; Perpendicular to x+y-2=0 (1,1), k=1, y=x;

The center of the circle is on y=x, x=2, y=2, the radius is (2,2), and the distance from (1,1).

So the equation for a circle (x-2) 2+(y-2) 2=2

Interpret the Qingming omen: The center of the circle is here.

The distance of the y-axis is equal to 2, x=2 or -2.

You may wish to have the center of the circle (2,b) and the radius r.

When the center of the circle is (2,b), the distance from the center of the circle (2,b) to the tangent line x+y-2=0 d=|﹣2＋b－2|／√2=r...

Circle c and straight line x+y-2=0

Tangent to p(1,1), r= [3 (b 1)]2 b 2|／√2=√

3²＋（b－1）²】

b－4）²=2【3²＋（b－1）²】

Simplified: b 4b 4 = 0

b＋2)²=0,b=﹣2

r=√3²＋（b－1）²】3√2

The equation for a circle is (x 2) y 2) =18

When the center of the circle is (2,b), the distance from the center of the circle (2,b) to the tangent line x+y-2=0 d=|2＋b－2|／√2

b|／√2r...

Circle c and straight line x+y-2=0

Tangent to p(1,1), r= [1 (b 1)] b|／√2=√

1²＋（b－1）²】

b²=2【1＋（b－1）²】

Simplified: b 4b 4 = 0

b－2)²=0,b=2

r=√1＋（b－1）²】2

The equation for a circle is (x 2) y 2) = 2 In summary, the equation for a circle is (x 2) y 2) =18 or (x 2) y 2) =2

The slope of the line l is k=1, then let the equation of the line l be: y=x+b, and the equation of the circle c is: x 2+y 2-2x+4y-4=0 then:

x 2+(x+b) 2-2x+4(x+b)-4=0, i.e.: 2x 2+2(b+1)x+b 2+4b-4=0so: x1+x2=-b a=-2(b+1) 2=-(b+1)x1*x2=c a=(b 2+4b-4) 2y1+y2=(x1+b)+(x2+b)=(x1+x2)+2b=-b-1+2b=b-1

Therefore, the midpoint of ab, i.e., the coordinates of the center of the circle is o(-(b+1) 2,(b-1) 2) again, ab= [(x1-x2) 2+(y1-y2) 2]= [(x1-x2) 2+(x1+b-x2-b) 2]= [2(x1-x2) 2].

=√[2*(-b^2-6b+9)]

Rather, ab=2r

So: r= [2*(-b 2-6b+9)] 2 Since the circle passes through the origin (0,0), then the distance between the center of the circle and the origin is equal to the radius of the circle, and the distance between the center of the circle and the origin d=

[(b^2+2b+1+b^2-2b+1)/4]=√[2*(b^2+1)]/2

So: [2*(-b 2-6b+9)] 2= [2*(b 2+1)] 2

Solution: b=1 or b=-4

Then, the equation for the straight line l is:

y=x+1 or, y=x-4

Let the equation for the circle be.

x-a)^2+(y-b)^2=r^2

After three points o(0,0),m(1,1),n(4,2)a2+b 2=r 2

a-1)^2+(b-1)^2=r^2

a-4)^2+(b-2)^2=r^2

The solution is a=4, b=-3, r=5

x-4) 2+(y+3) 2=25, the coordinates of the circle wheel are (including Tongqin 4,-3), and the radius r=5

Let the coordinates of the center of the circle be (a,5).

The distance from the center of the circle to the point (1,2) is equal to the radius.

So (a-1) +5-2) = 5

a²-2a-15=0

a+3)(a-5)=0

a=-3 or a=5

So the equation for a circle is (x+3) +y-5) =25 or (x-5) +y-5) =25

1.Let the equation for the circle be.

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

Crossing points a(0,0),b(1,1),c(4,2) so nuclear rock Kai.

f=01+1+d+e+f=0

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

Solution. d=-8,e=6,f=0

So. The circular equation is changed to x 2 + y 2-8x + 6y = 0 with the center of the circle being (4, -3) and radius = 16 + 9 = 5

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

The circular equation is (x-a) 2+(y-b) 2=25 because it passes through the point (1,2) and is tangent to the x-axis.

1-a)^2+(2-b)^2=25

b|=5, (obviously take b=5).

Solution. 1)a=-3,b=5;(2) a=5, b=5, the equation for the circle of the jujube liter is (x+3) 2+(y-5) 2=25 or (x-5) 2+(y-5) 2=25

Let the center of the circle cc(n,0), where n>0

The distance from the point c to the straight line is equal to the radius.

Listed: |3n+4|/5=2

i.e. n=2, or n=-14 3 rounded.

So the center of the circle is c(2,0).

That is, the equation for the circle c (x-2) +y = 4

2) Let the straight line l be y=kx+3

Get: (k +1) x + (6k-4) x + 9 = 0 so according to Vedder's theorem.

a+b=(-6k+4)/(k²+1),ab=9/(k²+1)bn=(12k+9)/(k²+1)

Since am+bn=3, =-1 or 5 is rounded off (there is no intersection with the circle), the equation for l is y=3-x, and the distance from the dot to the straight line is =3 2 2 The length of the chord gives 14

So according to the formula s=3 7 2