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3 years ago

✨ B R A I N L I E S T

Mathematics

2 answers:

nataly862011 [7]3 years ago

7 0

Step-by-step explanation:

3x^2+y^2-6x+6y-13=0

= 3x^2+y^2-6x+6y=13p

3x^2-6x+y^2+6y=13
3(x^2-2x)+y^2+6y=13
3(x^2-2x+?)+y^2+6y=13+?
3(x^2-2x+1)+y^2+6y=13+?
3(x^2-2x+1)+y^2+6y=13+3×1
3(x-1)^2+y^2+6y=16
3(x-1)^2+y^2+6y+?=16+?
3(x-1)^2+y^2+6y+9=16+?
3(x-1)^2+y^2+6y+9=16+9
3(x-1)^2+(y+3)^2=25
3(x-1)^2/25 + (y+3)^2/25 =1
(x-1)/ 25/3 + (y+3)^2/25 =1
(x-1)^2/ 25/3 + (y+3)^2/25 =1
Ellipse with centre (1,-3)

miv72 [106K]3 years ago

4 0

−6x−9y+13=0

(x−3)

+(y−

)

−9−

+13=0

(x−3)

+(y−

)

Here,

=(3,

)

Equation of another circle-

−2x−16y=0

(x−1)

+(y−8)

−1−64=0

(x−1)

+(y−8)

=65

Here,

=(1,8)

Distance between the centre of two circles-

(3−1)

+(8−

)

∣r

−r

∣=

∣

−

∣

∵C

=∣r

−r

∣

Thus the two circles touches each other internally.

Since the circle touches each other internally. The point of contact P divides C

externally in the ratio r

, i.e.,

=1:2

Therefore, coordinates of P are-

⎝

⎜

⎛

1−2

1(1)−2(3)

1−2

1(8)−2(

)

⎠

⎟

⎞

=(5,1)

Therefore,

Equation of common tangent is-

−S

(5x+y−6(

x+5

)−9(

y+1

)+13)−(5x+y−2(

x+5

)−16(

y+1

))=0

−6x−9y−13

+x+8y+13=0

4x−7y−13=0

Hence the point of contact is (5,1) and the equation of common tangent is 4x−7y−13=0.

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3 years ago

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Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

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frosja888 [35]

Answer:

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4 years ago