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It is known that the two roots of the bivariate linear equation of X: (6-k) (9-k) x ^ 2 - (117-15k) x + 54 = 0 are integers, and all k values that meet the conditions are obtained
:（6-k)(9-k)x^2-(117-15k)x+54=0
Factorization [(6-k) x - 9] [(9-k) x - 6] = 0
The solution is X1 = 9 / (6-k), X2 = 6 / (9-k)
Because the solution is an integer, K satisfying the condition is 3,7,15
It is known that the two roots of the quadratic equation (6-k) (9-k) X2 - (117-15k) x + 54 = 0 with respect to X are integers, and the values of all real numbers K satisfying the conditions are obtained
The original equation can be reduced to: [(6-k) X-9] [(9-k) X-6] = 0. Because this equation is a quadratic equation of one variable with respect to x, K ≠ 6, K ≠ 9, then there are: X1 = 96 − K (1), X2 = 69 − K (2). From (1) we get k = 6X1 − 9x1, from (2) we get k = 9x2 − 6x2, ≠ 6X1 − 9x1 = 9x2 − 6x2, So X1 = - 9, - 6, - 5, - 4, - 2, - 1, 0, 3. And K = 6X1 − 9x1 = 6-9x1. Substituting X1 = - 9, - 6, - 5, - 2, - 1, 3 respectively, we get k = 7152395334212, 15, 3
As shown in the figure, it is known that the two ends of the diameter AB passing through the circle O are the perpendicular lines am, BN and the perpendicular feet of the tangent line passing through the other point C on the circle O are m and N respectively
Why is C the midpoint of Mn
∵ am ⊥ Mn, BN ⊥ Mn, ∥ am ∥ BN, ∥ amnb are trapezoidal
∵ Mn cuts ⊙ o to C, ∵ OC ⊥ Mn, ∵ OC ∥ am, and AO = Bo, ∵ OC is the median line of trapezoidal amnb,
C is the midpoint of Mn
As shown in the figure: ⊙ O's diameter AB = 12, am and BN are its two tangent lines, de cut ⊙ o to e, intersect am to D, intersect BN to C, let ad = x, BC = y, find the functional relationship between Y and X, and draw its general image
If we make DF ⊥ CB through D and intersect CB at point F, ∵ DA and DC are tangent lines of circle O, ∵ Da = De, CB and CE are tangent lines of circle O, ∵ CB = CE, ∵ DAB = ∵ ABF = ∵ BFD = 90 °, ∵ quadrilateral ABFD is rectangle, ∵ Da = FB, DF = AB, in right triangle CDF, ∵ ad = x, BC = y, ab = 12, ∵ CD = CE + ed
As shown in the figure, the diameter of circle O AB = 12, am and BN are its two tangents, de cuts circle O to e, intersects am to D, intersects BM to C. let ad = x, BC = y, find the function of Y and X
y=36/x
x^2+y^2+6^2+6^2=(x+y)^2
2xy=2*6^2
xy=36
As shown in the figure: ⊙ O's diameter AB = 12, am and BN are its two tangent lines, de cut ⊙ o to e, intersect am to D, intersect BN to C, let ad = x, BC = y, find the functional relationship between Y and X, and draw its general image
Make DF ⊥ CB through D, intersect CB at point F, ∵ DA and DC are tangent lines of circle O, ∵ Da = De, CB and CE are tangent lines of circle O, ∵ CB = CE, DAB = ∵ ABF = ∵ BFD = 90 °, ∵ quadrilateral ABFD is rectangle, ∵ Da = FB, DF = AB, in right triangle CDF, ∵ ad = x, BC = y, ab = 12, ∵ CD = CE + ed = Da + CB = x + y, DF = AB = 12, CF = cb-fb = Y-X, according to Pythagorean theorem: CD2 = df2 + CF2, that is (x + y) ）2 = 122 + (Y-X) 2, the reduction is xy = 36, that is y = 36x (x > 0); draw the function image in the plane rectangular coordinate system, as shown in the figure
As shown in the figure: ⊙ O's diameter AB = 12, am and BN are its two tangent lines, de cut ⊙ o to e, intersect am to D, intersect BN to C, let ad = x, BC = y, find the functional relationship between Y and X, and draw its general image
Make DF ⊥ CB through D, intersect CB at point F, ∵ DA and DC are tangent lines of circle O, ∵ Da = De, CB and CE are tangent lines of circle O, ∵ CB = CE, DAB = ∵ ABF = ∵ BFD = 90 °, ∵ quadrilateral ABFD is rectangle, ∵ Da = FB, DF = AB, in right triangle CDF, ∵ ad = x, BC = y, ab = 12, ∵ CD = CE + ed = Da + CB = x + y, DF = AB = 12, CF = cb-fb = Y-X, according to Pythagorean theorem: CD2 = df2 + CF2, that is (x + y) ）2 = 122 + (Y-X) 2, the reduction is xy = 36, that is y = 36x (x > 0); draw the function image in the plane rectangular coordinate system, as shown in the figure
As shown in the figure, it is known that AB is the diameter of OD, am and BN are two tangents with ⊙ o, point E is the point on ⊙ o, and point D is the point on am
Connect de and extend intersection BN to point C, connect od and be, and OD ‖ be
Picture. Draw it yourself
The answer. Find it yourself
If you make mistakes in the title, you need someone else to draw for you
As shown in the figure, AB is known to be the diameter of ⊙ o, am and BN are its two tangent lines, CD and ⊙ o are tangent to point E, am and BN intersect at point C.D, AC = 4cm, BD = 9cm, (1) find the length of CD:
The topic is very confusing,
AB is diameter, am and BN are tangent,
∴AM∥BN,
Am and BN do not intersect,
Even if the letter pairs are reversed,
An function BN is tangent,
It is also impossible to have two intersections C and D
There is only one circle passing through points a (0,1) and B (4, m) and tangent to the x-axis. Find the value of M and the equation of the corresponding circle at this time
There are two situations
One is that the tangent point of the circle and the X axis coincides with B, that is, B is on the X axis. At this time, a vertical line of the X axis is made through B. the intersection of the vertical line and the middle vertical line of AB is the center of the circle. The two lines determine a point, so the center of the circle is unique, and the radius is equal to the distance from the center of the circle to the X axis
In this case, M = 0, let (x, y) be the center of the circle
Then B is the vertical line of X axis: x = 4 (1)
The perpendicular of AB: y = 4 (X-2) + 1 / 2 (2)
Simultaneous (1) (2) gives x = 4, y = 17 / 2, r = y = 17 / 2
So the equation of circle (x-4) ^ 2 + (y-17 / 2) 2 = (17 / 2) ^ 2
It is reduced to x ^ 2-8x + 16 + y ^ 2-17y = 0
Another case is that AB is parallel to the X axis. At this time, the vertical line of AB is perpendicular to the X axis, and the circle is tangent to the X axis, so the vertical line of AB passes through the tangent point. At this time, there is only one point with equal distance from the line to the three points, so the center of the circle is unique, and the radius is equal to the distance from the center of the circle to the X axis
Here M = 1
Let (x, y) be the center of the circle
This is the perpendicular of AB: x = 2
The radius is equal to the distance from the center of the circle to the x-axis and the distance from the center of the circle to a
So: y ^ 2 = 2 ^ 2 + (Y-1) ^ 2
Y = 5 / 2
So the equation of circle: (X-2) ^ 2 + (Y-5 / 2) ^ 2 = (5 / 2) ^ 2
It is reduced to x ^ 2-4x + 4 + y ^ 2-5y = 0
RELATED INFORMATIONS
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