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In the plane rectangular coordinate system, take point a (0,3) and point B (4,0) as the center of the circle, and take 8 and 3 as the radius to make circle a and circle B respectively, and find out the position relationship between the two circles
Center distance: D = √ (3 ^ 2 + 4 ^ 2) = 5
Radius relation: 8-3 = 5
So two circles are circumscribed
What is the position relationship between two circles 1 (0,3) and 2 (4,0) with radius of 8 and 3 in plane rectangular coordinate system?
Endonuclease
The distance between the centers of two circles is 5, which is exactly equal to the difference of the radii of the two circles
The endotangent condition is satisfied``
Take point a (3, 0) as the center of the circle and 5 as the radius to draw a circle, then the coordinates of the intersection point of the circle a and X axis are () A. （0，-2），（0，8） B. （-2，0），（8，0） C. （0，-8），（0，2） D. （-8，0），（2，0）
Because the center of the circle is on the x-axis, it intersects the x-axis by two points,
The ordinate of both points is 0,
∵ the radius of the circle is 5,
The abscissa of the two points is 3-5 = - 2, or 3 + 5 = 8
The coordinates of the two points are (- 2,0), (8,0)
Therefore, B
In the plane rectangular coordinate system, if a circle is drawn with point a (0, - 3) as the center and 5 as the radius, then the coordinates of the point where the circle intersects with the negative half axis of the Y axis is
（0,-8）
x²+（y+3)²=25
When x = 0
(y+3)²=25
Y + 3 = 5 or y + 3 = - 5
Y = 2 (rounding) or y = - 8
So the point is (0, - 8)
As shown in the figure, in a rectangular coordinate system, draw a circle with P (2,1) as the center and R as the radius When there are four common points between 0P and the coordinate axis, the point C (0, b) is a common point between the 0P and the y-axis, and the point a (m, 0) is a common point between the 0P and the x-axis ① When r = 2 root sign 2, find the value of M and B ② Try to deduce the formula of B expressed by algebraic expression containing m
Guo Dunyong replied:
① The equation of the circle is (X-2) 2 + (Y-1) 2 = R 2,
Substituting C (0, b) and point a (m, 0) into the equation of the circle, we can get,
（0－2）²+（b－1）²=R²,b ²－2 b+5= R² （1）
（m－2）²+（0－1）²=R²,m ²－4m+5= R² （2）
R=PC=PA,
R ²=（2－0）²+（1－b）²=（2－m）²+（1－0）²,
R 2 = B 2 - 2 B + 5 = m 2 - 4 m + 5 = (2 √ 2) 2 = 8, (equivalent to the former)
∴b ²－2 b－3=0,（b－3）（b+1）=0,b1=3,b2=－1,
m²－4m－3=0,m=2±√7,m1=4.6458,m2=－0.6458,
② From B 2 B + 5 = m 2 - 4 m + 5, B 2 B - M 2 + 4 M = 0,
∴b=1±（1/2）√[4+（m²－4m）²].
Draw ⊙ B with B (0,3) as the center and 6 as the radius to find the intersection coordinates of the circle and the coordinate axis Find the coordinates of the intersection point
10. (3 times root number 3,0); (negative 3 times root number 3,0)
y；（0,-3）；（0,9）
When a circle is drawn with point P (1,2) as the center and R as the radius, and there are exactly three intersections with the coordinate axis, what conditions does r satisfy
In two cases, the first one intersects both X and Y axes, but r = √ 5 is known by Pythagorean theorem
Second, it is tangent to the x-axis and intersects the y-axis (the origin is far away from the x-axis, so tangent to the y-axis cannot cross the x-axis)
R=2
Let the image of quadratic function f (x) = x 2 + 2x + B (x ∈ R) have three intersections with two coordinate axes in the plane rectangular coordinate system xoy, The circle passing through these three intersections is marked as C ① Find the value range of real number B; ② Find the equation of circle C; ③ Ask whether circle C passes through a fixed point (its coordinates are independent of B)? Please prove your conclusion Have detailed steps! Thank you!
（Ⅰ）f（x）=x2+2x+b
The axis of symmetry x0 = - B / 2A = - 1,
∵ there are two intersections with the X axis,
In the plane rectangular coordinate system xoy, it is noted that there are three intersections between the quadratic function f (x) = x2 + 2x + B (x ∈ R) and two coordinate axes. The circle passing through the three intersections is denoted as C (1) Find the value range of real number B; (2) Find the circle equation C; (3) Does circle C pass through a fixed point (its coordinates are independent of B)? Please prove your conclusion
(1) let x = 0, the intersection of parabola and Y axis is (0, b);
Let f (x) = x2 + 2x + B = 0. From the meaning B ≠ 0 and △ > 0, we can get B < 1 and B ≠ 0
(2) Let the general equation of the circle be x2 + Y2 + DX + ey + F = 0
Let y = 0 be x 2 + DX + F = 0, which is the same equation as x2 + 2x + B = 0, so d = 2, f = B
Let x = 0 be Y2 + ey + F = 0, and the equation has a root B. substituting it, we get e = - B-1
So the equation of circle C is x2 + Y2 + 2x - (B + 1) y + B = 0
(3) Circle C must pass through a fixed point, which is proved as follows:
Suppose that circle C passes through a fixed point (x0, Y0) (x0, Y0 does not depend on B), and the coordinates of this point are substituted into the equation of circle C,
The deformation is as follows: X02 + Y02 + 2x0-y0 + B (1-y0) = 0 (*)
In order to make the (*) formula hold for all B satisfying B ＜ 1 (B ≠ 0), there must be 1-y0 = 0. Combined with (*) formula, X02 + Y02 + 2x0-y0 = 0 is obtained
x0＝0
Y0 = 1 or
x0＝−2
y0＝1
It is found that (- 2,1) and (0,1) are on circle C. Therefore, circle C passes through fixed points (- 2,1) and (0,1)
In the plane rectangular coordinate system, let the image of quadratic function f (x) = x * x + 2x + B (x belongs to R) have three intersections with the two coordinate axes The circle passing through these two intersections is called C 1: Find the value range of real number B 2: Find the equation of circle C 3. Ask whether the circle C passes through a fixed point
Three intersections:
(0, b), (- 1 + radical (1-B), 0), (- 1-radical (1-B), 0)
1. Because the equation has a solution
∴b=<1
3. Let the center of the circle be (C, d) and the radius R
c=-1
∴1-b+d^2=r^2
1+(b-d)^2=r^2
∴b=2d-1
∵ the other intersection point of circle and Y axis is (2d-b, 0)
The circle passes through the fixed point (1,0)
2. Center of circle (- 1, (1 + b) / 2) R ^ 2 = (b ^ 2-2b + 5) / 4
The circle equation is (x + 1) ^ 2 + (y-0.5-0.5b) ^ 2 = 0.25b ^ 2-0.5b + 1.25
The result is: x ^ 2 + 2x + y ^ 2 - (B + 1) y + B ^ 2 = 0
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