High School Math Circle Equation Two Questions!Hurry!

Let the coordinates of m (x,y).

pm/ma=op/oa=1/2

Abscissa of point p = (3x-2) 2

The ordinate of point p = 3y 2

So the trajectory equation is (x-2 3) +y =4 9

Question 1: The equation can be set to x 2 + y 2 + ax + by + c = 0

1,2), (3,4) substitution:

a+2b+c+5=0 ……1）

3a+4b+c+25=0 ……2）

The chord length on the x-axis is 6, that is, the difference between the two roots of the quadratic equation about x is 6 when y=0, so that y=0:

x^2+ax+c=0

The relationship between the root and the coefficient: x1+x2=-a, x1x2=c, has: |x1-x2|=6

x1-x2|=√x1+x2)^2-4x1x2]=√a^2-4c)=6

i.e.: a 2-4c = 36 ......3）

Solve the system of ternary quadratic equations composed of (1), (2), and (3) to obtain:

a1=12，b1=-22，c1=27；

a2=-8，b2=-2，c2=7。

The equation for substituting the circle is as:

x 2 + y 2 + 12x-22y + 27 = 0, or x 2 + y 2 - 8x - 2y + 7 = 0

i.e., (x+6) 2+(y-11) 2=130, or (x-4) 2+(y-1) 2=10

Question 2: Introduce circular equations.

Let the circle be: x 2 + y 2 + 6 x - 4 + k (x 2 + y 2 + 6 y - 28) = 0....1)

x^2+y^2+6/（1+k)x+6k/(1+k)y+(-4-28k)/（1+k)=0

So the center of the circle is x=-3 (1+k) y=-3k (1+k).

Substituting x-y-4=0

Solve k = -7 and bring in (1) to get the equation obtained.

Give it points.

Isn't it (5a+1-1) 2+(12a) 2 1 169a 2 1

a^2＜1/169

1/13＜a＜1/13

The answer is right.

That is, the distance d from the point to the center of the circle is less than the radius.

So d is centered on (1,0) and r=1

So {5a+1-1) +12a) <1 25a +144a <1

a²-1/169<0

a+1/13)(a-1/13)<0

1/13

Solution: The equation for the circle: (x-1) +y = 1, the center of the circle (1,0), and the radius = 1.

The point (5a+1,12a) is in the circle, so the distance from the point to the center of the circle is less than 1. ∴√5a)²+12a)²]1.===>13|a|＜1.

=>|a|＜1/13.===>-1/13＜a＜1/13.

If you substitute x=5a+1 and y=12a into the circular equation less than or equal to 1, you get your result!As for what the answer is, it doesn't matter!

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

a+3b-26|Root 10 = r

2-a)^2+(-4-b)^2=r^2(8-a)^2+(6-b)^2=r^2

a+b=4 and then bring in the equation and you're good to go.

[This is an indefinite equation, and it is a problem to find its special solution. Solution: 27x-256y-175=0

=>27x=256y+175.<==>x=[27×9y+27×6+13y+13]/27=(9y+6)+[13(y+1)/27].Since x,y z+, and 13 is not divisible by 27, then y+1 must be divisible by 27, y+1=27t,(t z+)

i.e. y=27t-1When t=1, (y)min=26, at this time, x=253

Where did m,n come from?

Oh, and make the equation x=(256y+175) 27, and start algebra from 1.

Using the image, draw an image of this equation and look at the image.

The center of the circle (a, b), the circle of radius r, equation:

x-a) ten(y-b) r

Here, r=|ab|

r²=|ab|=(3-(-2)) Ten (-4-0) =25 Ten 16 41

Circular equation: x ten 2) ten y = 41

I know the center of the circle.,Wouldn't it be over if you set a standard formula and bring another point in.,The difficulty index is 0.。

The equation for a circle (x-2) +y-2) = 2 , the coordinates of point m are (a2, b 2).

The distance from the point m to the point (2,2) is 2

Then the trajectory equation for point m is (a 2-2) + b 2-2) = 2 (a>4,b>4).

Of course, you can use x, y to denote it.

I hope you can understand, and if you don't understand, you can ask.

Let m be (x,y).

The linear equation for ab is y=b a(x-x)+y

The distance from ab to (2,2) is 2 horizontally

Substitution to solve.

From the meaning of the title: the center of the circle c(0,4), let the line l and the circle tangent to the closed fibrillation p, then op=2, oc=4. Let the angle between the line l and the y axis be , then sin = op oc = 1 2.

30°。In this case, the angle between the linear L and the X-axis is 60°. Therefore, it is necessary to make the straight line l and the circle have two different intersection points.

Then k tan60° = 3, i.e. k 3.

①om∣=√x₁²+kx₁²)x₁√(1+k)，∣on∣=√x₂²+kx₂²)x₂√(1+k)。

2 m = 1 x +1 x = 2k n, i.e. n = 2k (x x ) x +x ) 2x x]. Now we substitute y= k·x into the circular equation.

1+k)x²-8√k·x+12=0。So x +x = 8 k (1+k) and x x = 12 (1+k). Substituting gives n = 36k (5k-3) = 36 (5-3 k).

Because k 3, so 6 5 5 n 3. Key Mountain.