The intersection point is only one. Then the equation of the line is tangent to the circle at point (1, -2).
<h3>What is a circle?</h3>
It is a locus of a point drawn an equidistant from the center. The distance from the center to the circumference is called the radius of the circle.
Prove algebraically that the straight line with equation x = 2y + 5 is a tangent to the circle with equation x² + y² = 5.
x = 2y + 5 ...1
x² + y² = 5 ...2
If the intersection of the point of the circle and line is one. Then the line is tangent to the circle.
Then from equations 1 and 2, we have
(2y + 5)² + y² = 5
4y² + 25 + 20y + y² - 5 = 0
5y² + 20y + 20 = 0
5y² + 10y + 10y + 20 = 0
5y (y + 2) + 10(y + 2) = 0
(5y + 10)(y + 2) = 0
y = -2, -2
Then the value of y is unique then the value of x will be unique.
The value of x will be
x = 2(-2) + 5
x = -4 + 5
x = 1
The intersection point is only one. Then the equation of the line is tangent to the circle at point (1, -2).
More about the circle link is given below.
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....................................do you mean 3.11%
Yes it is true:
(1 - cos^2 x) = sin^2 x Sqr.(1 - cos^2 x) = sin x and (-sinx) True, since sin x negative. Also, x terminates in quadrant 3 and 4. Hope this is helpful
This figure is a parallelogram because the opposite sides are congruent. Side CG would become 28 and side BH would also solve out to be 28
T=-1
sinA=sin(π/2-3A), A=2nπ+π/2-3A, 4A=2nπ+π/2, A=nπ/2+π/8 where n is an integer.
Also, π-A=2nπ+π/2-3A, 2A=2nπ-π/2, A=nπ-π/4.
The hard way:
cos3A=cos(2A+A)=cos(2A)cosA-sin(2A)sinA.
Let s=sinA and c=cosA, then s²+c²=1.
cos3A=(2c²-1)c-2c(1-c²)=c(4c²-3).
s=c(4c²-3) is the original equation.
Let t=tanA=s/c, then c²=1/(1+t²).
t=4c²-3=4/(1+t²)-3=(4-3-3t²)/(1+t²)=(1-3t²)/(1+t²).
So t+t³=1-3t², t³+3t²+t-1=0=(t+1)(t²+2t-1).
So t=-1 is a solution.
t²+2t-1=0 is a solution, t²+2t+1-1-1=0=(t+1)²-2, so t=-1+√2 and t=-1-√2 are solutions.
Therefore tanA=-1, -1+√2, -1-√2 are the three solutions from which:
A=-π/4, π/8, -3π/8 radians and these values +2πn where n is an integer.
Replacing π by 180° converts the solutions to degrees.