Find the equation of the circle, find the general equation of the circle

Let the equation for the circle be:

x-a) 2+(y-b) 2=c 2(c>0) round c: (x-1) 2+y 2=1

From the nature of the excision:

a-1) 2+b 2=(c+1) 2 (1) circle through the point (3,- 3), substituting :

3-a) 2+(-3-b) 2=c 2 (2) The straight line is tangent to the circle, and the straight line passing through the point (a, b), (3, 3) is perpendicular to the straight line x+ 3y=0.

b+ 3) (a-3)*(1 3)=-1 (3) is obtained by equation (3).

b+√3=√3(a-3)

Both sides are squared.

b+√3)^2=3(a-3)^2

Substituting equation (2) obtains:

4(a-3)^2=c^2

c=|2a-6|

Obtained from (3).

b=√3a-4√3

When c=2a-6, substituting equation (1) obtains.

a=4 c=2a-6=2>0

b=0 When c=6-2a, substituting equation (1) obtains.

a=0 c=6-2a=6>0

b=-4√3

So the equation for the circle is.

x-4) 2+y 2=4 or x 2+(y+4 3) 2=36 so many people answered, sure enough, there will be brave men under the heavy reward! However, the above few are right, and the answer to the pending coefficient method is not right.

Let the equation for the circle c: (x-a) 2+(y-b) 2=r 2

The circle c is tangent to the circle x 2 + y 2-2x = 0, i.e. (x-1) 2 + y 2 = 1, then.

The distance between the two centers of the circle = the sum of the two radii.

a-1)^2+b^2]=r+1...1)

Tangent to the line x+ 3y at the point q (3,- 3), indicating that the point q is on the circle c, that is, satisfying:

3-a)^2+(-3-b)^2=r^2...2)

and the distance from the center of the circle to the tangent = r

The distance from the center of the circle x 2 + y 2 - 2x = 0 to the straight line x + 3y = 0: 1 2

So there is: (1 2) r=1 (r+1), we get r=1, and we get the solution by substituting (1) and (2).

a=2, b=-3, or a=20 7, b=-3 3 7

So the equation for a circle: (x-2) 2+(y+ 3) 2=1 or (x-20 7) 2+(y+3 3 7) 2=1

The garden equation is. x-a) 2 + y-b) 2 = c 2 then there is a system of equations:

3-a)^2 + 3-b)^2 = c^2c= | a + 3b|/2

1-a) 2 + 0-b) 2 = (1+c) 2 Solve it yourself

Because the circle c is tangent to the straight line x+ 3y=0 at the point (3,- 3).

Then the straight line through the point (3,- 3) and the center of the circle must be perpendicular to the straight line x+ 3y=0

The slope of this line is k = root number 3

The equation is y+3=root number 3 (x-3).

The center of the circle is in this straight line.

Let the center of the circle be o(a, root number 3*a-4 root number 3).

Straight line x + 3y = 0

r=d(distance from the center of the circle to the straight line)=|a+3 (root number 3*a-4 root number 3)|/2=|2a-6|

Because x 2 + y 2 - 2x = 0

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

c(1,0)

and oc 2 = (1 + r) 2 = (a-1) 2 + (root number 3 * a-4 root number 3) 2

Let a>=3, (2a-5) 2=(a-1) 2+(root number 3*a-4 root number 3) 2

a=4 if a<3 , 7-2a) 2=(a-1) 2+(root number 3*a-4 root number 3) 2

a=0 so o(4,0) r=2 or o(0,-4 root number 3) r=6

The equation for a circle is (x-4) 2+y 2=4 or x 2+(y+4, root number 3) 2=36

If the circle is to be tangent to the line x+ 3y=0 at (3,- 3), then the line at the center of the circle is y= 3x-4 3

Set the coordinates of the center of the circle (x, 3x-4 3).

then the Pythagorean theorem (x-1) 2+( 3x-4 3) 2=(2|).x-3|+1)^2

2|x-3|+1 is the distance between the circle c and the circle sought The distance from the center of the circle is also the hypotenuse of the triangle divided into x>3 and x<3.

Solve for x=4 or x=0

The radius of the center coordinate (4,0) is equal to (3,0) and the distance from (3,- 3) is equal to 2

Then the original equation is (x-4) 2 + y 2 =4, the coordinates of the center of the circle (4,-4 3 ) and the radius is (4,0) and (4,-4 3 ) The distance is equal to 6

Then the original equation is x 2 + y+4 3 ) 2=36 Friend, I need this score I believe mine can also satisfy you, do me a favor.

The direct method lists the equations from the geometric conditions that the moving points given by the problem are satisfied, and then substitutes and simplifies the coordinates to obtain the trajectory equation obtained, which is called the direct method.

Example 1 Knowing that the sum of the distances from the moving point p to the fixed point f(1,0) and the straight line x=3 is equal to 4, find the trajectory equation for the point p.

Solution: If the coordinates of the point p are (x,y), then it can be obtained from the title.

1) When x 3, the equation becomes , simplified to .

2) When x3, the equation becomes , simplified to .

Therefore, the trajectory equation for the point p is or .

2. Definitions.

From the geometric conditions satisfied by the moving point given by the problem, after simplification and deformation, it can be seen that the moving point satisfies the definition of the quadratic curve, and then the trajectory equation is obtained, which is called the definition method.

Example 2 Knowing that the center of the circle is m1 and the center of the circle is m2, a moving circle and these two circles are tangential to find the trajectory equation of the center of the circle p.

Solution: Let the radius of the moving circle be r, which can be obtained from the condition of the two circles tangent: , the trajectory of the center of the moving circle p is the right branch of the hyperbola focusing on m1 and m2, c=4, a=2, b2=12.

Therefore, the trajectory equation is .

3. Pending coefficient method.

The type of curve can be known from the meaning of the question, the equation is set into the general form of the curve equation, and the required undetermined coefficient is obtained by using the conditions given by the problem, and then the trajectory equation is obtained, which is called the undetermined coefficient method.

Example 3 Knowing that the center of the hyperbola is at the origin and a focal point is f( ,0), the line y=x 1 intersects it at m and n points, and the abscissa of the midpoint of mn is , the hyperbolic equation is obtained.

Derived from Veda's theorem. There is also , a system of simultaneous equations, which can be solved.

The equation for this hyperbola is .

Fourth, the parametric method.

Select the appropriate parameters, use the parameters to represent the coordinates of the moving point, obtain the parameter equation of the moving point trajectory, and then eliminate the parameters to obtain the ordinary equation of the moving point trajectory, this method is called the parameter method.

Example 4 Crossing the origin point as a straight line l and a parabola intersect at two points A and B, and find the trajectory equation for the midpoint M of the line segment AB.

Solution: From the analysis of the meaning of the problem, we know that the slope of the straight line l must exist, and let the equation of the straight line l y=kx. Substituting it into a parabolic equation gives . Because the straight line and the parabola intersect, 0 is solved.

Let a( )b( )m(x,y), which is obtained by Veda's theorem.

Obtained by eliminating k.

Again, so .

The trajectory equation for point m is.

I only have these four, and it's enough to cope with high school math.

Solution: From the meaning of the question, the coordinates of the center of the circle can be set as (3a, a).

Since the circle is tangent to the y-axis, so:

The radius r=|3a|

Again, the straight line y=x, that is, x-y=0, the chord length obtained by the truncation is 2 7, and the distance from the center of the circle (3a, a) to the straight line x-y=0 d=|3a-a|/√2=√2*|a|

Then the vertical diameter theorem can be obtained:

3a|²=(√7)²+2*|a|) i.e. 9a = 7+2a

7a = 7 gives a = 1 or a = -1

Then when a=1, the center of the circle is (3,1), the radius r=3, and the equation for the circle is (x-3) +y-1) =9

When a=-1, the center of the circle is (-3,-1), the radius r=3, and the equation for the circle is (x+3) +y+1) =9

Let the square of the circle (x-a) +y-b) = r The circle c and the y-axis are tangent |a|=r The center of the circle is on the line x-3y=0, a-3b=0 because the chord length truncated by the line y=x is 2 7, r -a -2ab+b 2=7 concatenate these three equations to find the circle equation as (x-3) +y-1) =9 or (x+3) +y+1) =9

The center of the circle is (2,0), the radius is the root 10, and the equation for the circle is (-2)2+y2=10

The intersection of the tangent of the center of the circle at the crossing point b and the perpendicular bisector of ab is the perpendicular bisector of ab: (x-3) 2+(y-1) 2=(x-1) 2+(y-2) 2

That is, 4x-2y-5=0

The tangent of the crossing point b: k=, b=y=

The center of the circle x=2,,,y=

So the circular equation is:

x-2)^2+(=

Since the circle is tangent to the straight line 2x-y=0

Therefore, it can be seen that the slope of the straight line at the center of the circle is -1 2

Let the equation for the straight line through the center of the circle be y=kx+b

Bringing point b in gives b=5 2

So the equation for the center of the circle is x+2y=5

Let the center of the circle be (5-2a,a).

The distance from the center of the circle to any point on the circle is equal.

So, the point-to-straight distance formula.

d = under the root number (5-2a-1) 2 + (a-2) 2d = under the root number (5-2a-3) 2 + (a-1) 2 solution to obtain a = 3 2

So the center of the circle is (2,3 2).

The solution gives d = 5 2, which is the radius.

So the equation for a circle is (x-2) 2+(y-3 2) 2=5 4

In the standard equation of the circle (x-a) +y-b) =r, there are three parameters a, b, r, that is, the coordinates of the center of the circle are (a, b), only a, b, r are required, then the equation of the circle is determined, so to determine the circle equation, there must be three independent conditions, where the center coordinate is the positioning condition of the circle, and the radius is the shaping condition of the circle.

It is easy to determine the center and radius of the circle.

It is known that the chord length obtained by the circle satisfies the truncated y-axis is 2 and is divided into two arcs by the x-axis, and the ratio of the arc length is 3 :1 The distance from the center of the circle to the straight line l: x-2y=0 is 5 5, and the equation for the circle is found.

Solution: Let the equation for the circle be (x-a) +y-b) =r .1)

The distance from the center of the circle m(a,b) to the straight line x-2y=0 is 5 5, so there is the equation:

a-2b 5 = 5 5, therefore.

a-2b=-1...2)

or a-2b=1....3)

Let the intersection points of the circle and the y-axis be (0,y1) and (0,y2), and substitute x=0 into (1) to obtain:

y²-2by+a²+b²-r²=0

Because "the chord length obtained by the truncated y-axis is 2", i.e. y1-y2 =2According to Veda's theorem, there is an equation:

y1-y2)²=(y1+y2)²-4y1*y2

4b²-4(a²+b²-r²)

4(r²-a²)=4

So we get: r -a =1....4)

It is also "divided into two arcs by the x-axis, and the ratio of the arc length is 3:1", and the center of the circle is set to the inferior arc s1.

The angle is 1, and the central angle of the circle to which the optimal arc s2 pairs is 2, then.

s2 s1=r 2 r 1= 2 1=3 1, so 1=90 , 2=270

Let the arc intersect with the x-axis at two points, a and b, then amb is an isosceles right triangle, and therefore a chord.

Long ab = x1-x2 = (2)r

Let y=0 in Eq. (1), we get:

x²-2ax+a²+b²-r²=0

Thus by the Vedic theorem there is:

x1-x2)²=(x1+x2)²-4x1*x2=4a²-4(a²+b²-r²)

4(r²-b²)=2r²

i.e. r -2b = 0....5)

From (2), (4) and (5), the simultaneous solution is obtained: a=1, b=1, r =2

The equation for the circle is: (x-1) +y-1) =2

From (3), (4) and (5), the simultaneous solution is obtained: a=-1, b=-1, r =2

The equation for the circle is: (x-1) +y-1) =2

Solution: Let the midpoint of ab be m and the radius of circle c r

The focus of the parabola y 2=4x is (1,0).

Because the center of the circle c is symmetrical with the focus of the parabola y 2=4x with respect to the straight line y=x.

Then the center of the circle c is (0,1).

That is, the distance from the center of the circle c to the straight line 4x-3y-2=0 d=1 because |ab|=6

Then am=3 is obtained by the Pythagorean theorem r 2=am 2 + d 2

The equation for calculating the circle c is: x 2+(y-1) 2=10