The equation of circle: given the points P1 (4,9), P2 (6,3), circle C is a circle with the diameter of line p1p2, then the equation of circle C is the same Please write down the detailed process, thank you

Because circle C is a circle with the diameter of segment p1p2,
So the center of the circle is the midpoint of p1p2,
Namely (5,6)
The diameter is | p1p1 | = 40 under the root sign
So the equation
（x-5）^2+（y-6）^2=10

If two points P1 (4,9) and P2 (6,3) are known, then the equation for a circle with diameter p1p2 is______ ．

Let the midpoint of line p1p2 be m, ∵ P1 (4,9) and P2 (6,3), the center of circle m (5,6), and | p1p2 | = (4 − 6) 2 + (9 − 3) 2 = 210, and the radius of circle 12 | p1p2 | = 10, then the equation of circle is: (X-5) 2 + (y-6) 2 = 10. So the answer is: (X-5) 2 + (y-6) 2 = 10

Given that points P1 = (4,9) and P2 = (6,3) 0 are circles with the diameter of line segment p1p2, what are the following points on 0?
A (6,9) B (3,3) C (5,3) d (2,4)

The center coordinate of the circle is the midpoint of p1p2, i.e. (5,6)
The radius is the midpoint and the distance from P1 is the root 10
So the equation is (X-5) ^ 2 + (X-6) ^ 2 = 10
Over point a

Given the hyperbola x &# 178; - y &# 178 / 2 = 1, the straight line passing through point P (2,1) intersects the hyperbola at P1 and P2, and the trajectory equation of the midpoint m of line p1p2 is obtained

Let the linear equation over P (2,1) be Y-1 = K (X-2), that is, y = kx-2k + 1
The simultaneous hyperbola x ^ - y ^ / 2 = 1 and the analytical formula of this line, by eliminating y, we can obtain the quadratic equation of one variable about X
(k^-2)x^ - (4k^-2k)x +(4k^-4k+3)=0
And △ = (4K ^ - 2K) ^ - 4 (k ^ - 2) * (4K ^ - 4K + 3) = 24 (k - 2 / 3) ^ + 40 / 3 > 0
Let P1 (x1, Y1), P2 (X2, Y2) be the intersection points of straight line and hyperbola
Then the two different real roots of the above equation must be the abscissa x1, X2 of the two different intersections P1, P2 of the straight line and the hyperbola
x1+x2=(4k^-2k)/(k^-2) ①
The ordinates of P and Q are expressed by their abscissa respectively
y1=kx1-2k+1
y2=kx2-2k+1
∴y1+y2=k(x1+x2)-4k+2
By substituting formula (1), we get the following result:
y1+y2=(8k-4)/(k^-2) ②
According to the formula of midpoint coordinates, the coordinates of M (x, y) in p1p2 can be obtained as follows:
x=(x1+x2)/2
y=(y1+y2)/2
The results are as follows
x=(2k^-k)/(k^-2)
y=(4k-2)/(k^-2) ③
By comparing the two formulas, it is concluded that:
x/y=k/2
k=2x/y
By substituting this formula into (3), we get the following results
(x-1)^/(7/8) - (y-1/2)/(7/4) =1
(in the process of simplification, y is deleted from both sides of the equation, because it can be seen from the image that y cannot always be zero.)
That is, the trajectory of P is a hyperbola with the center at (1,1 / 2) and the intersection on the x-axis

Given hyperbola x ∧ 2-y ^ 2 / 2 = 1 (1), the line L passing through point a (2,1) intersects with the given hyperbola at P1 and P2, the trajectory equation of the midpoint P of line p1p2 is obtained. (2) whether the line L 'passing through point B (1,1) intersects with the given hyperbola at two points Q1 and Q2, and B is the midpoint of line q1q2, explain the reason

I'm doing it

The straight line passing through the left focus F of hyperbola x ^ 2 / 4-y ^ 2 = 1 intersects the hyperbola at P1 and P2. If p1p2 = 4, how many such straight lines are there?

X & sup2 / / 4-y & sup2; / = 1 = = > A = 2 ∵ distance between two vertices = 2A = 4, two points of p1p2 are just on the left and right half of the hyperbola, only one (the rest are all > 4) C = √ (4 + 1) = √ 5 = = = = > left focus f (- √ 5,0) if the straight line p1p2 ⊥ X axis, substitute x = - √ 5 into X & sup2 / / 4-y & sup2; = 1 = = = = > y = ± 1 / 2, | p1p2

The line m passes through the point a (- 4, - 2), and the point a is the midpoint of the line m cut by the two coordinate axes

Let m be x / A + Y / b = 1
Then the intersection point (0, b), (a, 0) with the coordinate axis
A is the midpoint
So (0 + a) / 2 = - 4, (B + 0) / 2 = - 2
a=-8,b=-4
-x/8-y*4=1
x+2y+8=0

Given that the point (4, 2) is the midpoint of the line segment cut by the ellipse x236 + Y29 = 1, then the equation of the line L is ()
A. x-2y=0B. x+2y-4=0C. 2x+3y+4=0D. x+2y-8=0

There are two points a (x1, Y1), B (X2, Y2), a (x1, Y1), B (X2, Y22) where a line L and ellipse intersect at two points a (x1, Y1, Y1), a (x1, Y1, Y1), B (X2, Y1), B (X2, Y1), B (X2, Y22), B (X2, Y22), B (X2, Y22). To the elliptic equation, you can get x2136 + y2136 + y219 = 1, x2236 + x2236 + y2236 + y2236 + y229 = 1 = 1, x2236 + y2236 + y229 = 1, two formula subtraction (x1 + x2) (x1 + x2) (x1 + x2) (x1 + x2) (x1 + x2) (x1 + x2) (x1 + x2136 + y2136 = 1 = 1 = 1, x2136 = 1, x2136 = 1, x2136 = 1, x2236 = 1, x2236 = 1, x2236 + x2236 + x2236 + x2236 + x2d

Given that the line L passes through the point P (3, - 1), the intersection of the line L and the coordinate axis at points a, B and P is the midpoint of the line AB, the equation of the line L is obtained
Why multiply by two?

According to the meaning of the title:
The coordinates of two points AB can be set as:
A (m, 0), B (0, n)
(m+0)/2=3;
(0+n)/2=-1;
So: M = 6, n = - 2
m. According to the intercept equation of the straight line, the equation of the straight line can be obtained as follows:
X / 6 + Y / (- 2) = 1
x-3y-6=0.

The line L passes through the point a (- 3,2), and the point a is the midpoint of the line segment obtained by the intersection of L and the coordinate axis. The equation of the line L and the intersection of the line L and the two coordinate axes are obtained

Let the equation of l be y = K (x + 3) + 2
When x = 0, y = 3K + 2
When y = 0, x = - 2 / K-3
Therefore, the midpoint coordinate of the intersection point with the coordinate axis is (- 1 / K-3 / 2,3k / 2 + 1)
So there are:
-1/k-3/2=-3 1）
3K / 2 + 1 = 2, from which k = 2 / 3,
Substituting the formula 1,
So the straight line is y = 2 / 3 * (x + 3) + 2 = 2x / 3 + 4
The intersection point with the coordinate axis is (0,4), (- 6,0)

The equation of circle: given the points P1 (4,9), P2 (6,3), circle C is a circle with the diameter of line p1p2, then the equation of circle C is the same Please write down the detailed process, thank you