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Deffense [45]
1 year ago
7

2. Consider the circle x² + y2 = 1, given in figure. Let OP makes an angle 30° with the x axis.

Mathematics
1 answer:
PolarNik [594]1 year ago
6 0

The equation of the tangent line to the circle passing through the point P is; y = (1/√3)x ± 2/√3

<h3>How to find the equation of the tangent?</h3>

I) We are given the equation of the circle as;

x² + y² = 1

Since angle of inclination is 30°, then slope is;

m = tan 30 = 1/√3

Then equation of the tangent will be;

y = (1/√3)x + c

Put  (√3)x + c into the given circle equation to get;

x² + ((1/√3)x + c)² = 1

x² + ¹/₃x² +  (2/√3)cx + c² = 1

⁴/₃x² +  (2/√3)x + (c² - 1) = 0

Since we need to find value of c for equation to become tangent, then the above quadratic equation must have real and equal roots.

Thus;

((2/√3)c)² - 4(⁴/₃)(c² - 1) = 0

⁴/₃c² - ¹⁶/₃(c² - 1) = 0

⁴/₃c² - ¹⁶/₃c² + ¹⁶/₃ = 0

4c² = ¹⁶/₃

c² = ⁴/₃

c = √⁴/₃

c = ±²/√3

Thus, equation of tangent is;

y = (1/√3)x ± 2/√3

II) Radius from the given equation is 1. Thus, we will use trigonometric ratio to find the x and y intercept;

x-intercept is at y = 0;

0 = (1/√3)x ± 2/√3

-(1/√3)x = ±2/√3

Intercept is positive. Thus;

x = (2/√3)/(1/√3)

x = 1

y - intercept is positive at x = 0;

y = (1/√3)0 ± 2/√3

y = 2/√3

Read more about Equation of tangent at; brainly.com/question/17040970

#SPJ1

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Match each set of vertices with the type of triangle they form.
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Answer:  The calculations are done below.


Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

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