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It is known that Tan α and Tan β are two solutions of the equation 2x square + 4x + 1 = 0
∵
tan(a+b)=(tana+tanb)/(1-tanatanb)
According to Weida's theorem
tana+tanb=-b/a=-2
tanatanb=c/a=1/2
∴
tan（a+b)=(-2)/(1-1/2)=-4
If a is the equation - x square + 2x + 15 = 0, prove that the equation 4x square - (A-1) x + 1 = O has two equal real roots
Substitute a into - x ^ 2 + 2x + 15 = 0 to get:
-a^2+2a+15=0
Namely:
a^2-2a-15=0
(a+3)(a-5)=0
A = - 3 or 5
By substituting - 3 and 5 into 4x squared - (A-1) x + 1 = 0, we obtain:
4X ^ + 4x + 1,4x ^ - 4x + 1
4X ^ + 4x + 1 has two equal real roots negative 1 / 2, 4x ^ - 4x + 1 has two equal real roots 1 / 2
Solving the equations y + x = 5 2x + y = k 2x + 3Y = 3k-4
Solving equations
y+x=5
2x+y=k
2x+3y=3k-4
2x+（5－x）＝k x+5＝k
2X + 3 (5-x) = 3k-4; reduced to 15-x = 3k-4
15-x = 3 (x + 5) - 4, so x = 1; k = 6; y = 4
If the solution of the system of equations 2x + 3Y = K + 4 (1) X-Y = 3k-3 (2) satisfies X-Y = 5, the value of K is obtained
To be more detailed process ah, thank you~~
Sorry, it's x + y = 5
If X-Y = 5, we can get that x = y + 5 is brought into the second formula and K = 8 / 3
Given that there is only one circle passing through points a (0, 1), B (4, a) and tangent to X axis, the value of a and the equation of the corresponding circle are obtained
Consider two cases: (I) let the center coordinate of the circle be (x, y), when point B is the tangent point, B is on the X axis, so a = 0. Then B (4, 0), so the midpoint coordinate of AB is (2, 12), and the slope of the line AB is 1 − 00 − 4 = - 14, then the slope of the vertical line AB is 4, so the equation of the vertical line AB is Y-12 = 4 (X-2
It is known that there is only one circle passing through a (0,1) and point B (4, a) and tangent to the x-axis. The value of a and the equation of the corresponding circle are obtained
Let the center of the circle be (x, y) x ^ 2 + (Y-1) ^ 2 = y ^ 2 （1）(x-4)^2+(y-a)^2=y^2 …… (2) By solving the equations, we can get x ^ 2-2y + 1 = 0 （1）x^2-8x+16-2ay+a^2=0 …… (2) Substituting 1 into 2, we can get (1-A) x ^ 2-8x + 16-a + A ^ 2 = 0, because there is an intersection point, so that △ = 0 to calculate a = 0, there are two
A circle passes through two points a (4,2), B (- 1,3), and the sum of the four intercepts on the two coordinate axes is 2
Let the equation of the circle be x2 + Y2 + DX + ey + F = 0
Let y = 0 get x2 + DX + F = 0,
The sum of intercept of circle on X axis is X1 + x2 = - D,
Let x = 0 give Y2 + ey + F = 0,
The sum of the intercepts of a circle on the y-axis is Y1 + y2 = - E,
Let X1 + x2 + Y1 + y2 = - (D + e) = 2,
∴D+E=-2①
And a (4,2), B (- 1,3) are on the circle,
∴16+4+4D+2E+F=0,②
1+9-D+3E+F=0,③
D = - 2, e = 0, f = - 12
So the equation of circle is: x2 + y2-2x-12 = 0
x2+y2-2x-12=0
Let y = 0 get x2 + DX + F = 0,
The sum of intercept of circle on X axis is X1 + x2 = - D,
Let x = 0 give Y2 + ey + F = 0,
The sum of the intercepts of the circle on the y-axis is Y1 + y2 = - E
How did you get here!
The intercept of a circle on the x-axis is the sum of the abscissa of the two intersections of the circle and the x-axis
That is to say, let y = 0 get the sum of intercept X1 + x2 of quadratic equation with one variable x2 + DX + F = 0 about X
According to the relationship between root and coefficient, there is X1 + x2 = - D
In the same way
Let x = 0 give Y2 + ey + F = 0,
The sum of the intercepts of the circle on the y-axis is Y1 + y2 = - E
A circle passes through two points a (4,2) B (- 1,3), and the sum of the four intercepts on the two axes is 4
Note that the sum of intercepts is 4, not 2
Let X & # 178; + Y & # 178; + DX + ey + F = 0
According to the meaning: X1 + x2 + Y1 + y2 = - D-E = 4,4d + 2E + F = - 20 - D + 3E + F = - 10
The solution is: D = - 7 / 3, e = - 5 / 3, f = - 22 / 3
The equation of circle is: X & # 178; + Y & # 178; - 7 / 3x-5 / 3y-22 / 3 = 0
A circle passes through two points a (4,2), B (- 1,3), and the sum of four intercepts on two coordinate axes is 2
Let the equation of circle be x2 + DX + Y2 + ey + F = 0, and substitute two points a (4, 2) and B (- 1, 3) into the equation to get: e = 5D + 10, f = - 14d-40, because the four intercepts are 2, so - D-E = 2, so the solution is: D = - 2, f = - 12, e = 0, so the equation of circle is x2-2x + y2-12 = 0, that is, (x-1) 2 + y2 = 13
Given that the fixed point a (a, 2) is outside the circle x ^ 2 + y ^ 2-2ax-3y + A ^ 2 + a = 0, find the value range of A
Kobe XXX,
x^2+y^2-2ax-3y+a^2+a=0
After finishing, we get
(x-a) ^ 2 + (Y-3 / 2) ^ 2 = 9 / 4 - a the center of the circle is: (a, 3 / 2)
So 9 / 4-A > 0 A2
Comprehensive formula [1]
9/4>a>2
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