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Answer:

#6# units

Explanation:

enter image source here
A circle with a radius #r# lying in the first quadrant and touching the coordinate axes has a center #C=(r,r)#, as shown in the figure.
Recall that the distance between a point #P(x_0,y_0)# and a line #L: ax+by+c=0# is :
#d=|ax_0+by_0+c|/sqrt(a^2+b^2)#
Given that the circle also touches the line #4x+3y-12=0#,
#=># the distance #r# from point #C(r,r)# to line #4x+3y-12=0# is :
#r=|4r+3r-12|/sqrt(4^2+3^2)#
#=> r=|7r-12|/5, => r=1 or 6# units,

Hence, the largest radius #r=6# units

enter image source here
Transforming the given equation of the circle #3x²+3y²+12x+8y+8=0# in standard form we get.

#3x²+3y²+12x+8y+8=0#

#=>x²+y²+4x+8/3y+8/3=0#

#=>(x²+2*x*2+ 2^2)+y^2+2*y*4/3+(4/3)^2=4+16/9-8/3#

#=>(x+2)^2+(y+4/3)^2=(sqrt(28/9))^2#

So the center of the circle C #to(-2,-4/3)#

and radius of the circle #rtosqrt(28/9)#

If #PandQ# are points of contact of two tangents drawn from origin #O# to the circle then #DeltasPOCandQOC# will be two congruent right triangles both having common hypotenuse OC.

So #PC=QC=r#

Now #OC^2=(0-(-2))^2+(0-(-4/3))^2#

#=>OC=sqrt(52/9)#

So length of each tangent

#OP=sqrt(OC^2-CP^2)#

#=sqrt(52/9-28/9)#

#=sqrt(8/3)=2sqrt(2/3)=(2sqrt6)/3#

Now #sinanglePOC=(PC)/(OC)#

#=>sinanglePOC=sqrt(28/9)/sqrt(52/9)#

#=>anglePOC=sin^-1(sqrt(7/13))=47.2^@#

So angle between the tangents will be

#anglePOQ=2*anglePOC=94.4^@#

But #angle PCQ=2*angle PCO=2*(90^@-47.2^@)=85.6^@#

Answer:

# d = 1.496 times 10^{11} cdot 1920/3600 cdot {2 pi}/360 = 1.39254×10^9 # meters

Explanation:

Well now I have to look up what the double quote means. Probably minutes or seconds.

OK, Wikipedia says For example, 40.1875° = 40° 11' 15" .

That's 40 degrees, 11 minutes, 15 seconds, a minute being 1/60th of a degree and a second being 1/60th of a minute. (Thank you Babylonians.)

So we're to understand 1920" as

# 1920/3600 # degrees # = 1920/3600 cdot {2 pi}/360 # radians

So we're looking at a diameter

# d = 1.496 times 10^{11} cdot 1920/3600 cdot {2 pi}/360 # meters

# d = 1.39254×10^9 # meters

Check: Google #quad sqrt#

enter image source here

Answer:

#color(blue)(4-sqrt(10)color(white)(888)"units"#

Explanation:

The general equation of a circle is:

#(x-h)^2+(y-k)^2=r^2#

Where:

#bbh# and #bbk# are the #x# and #y# coordinates of the centre respectively, and #bbr# is the radius.

Circle A

#(x-3)^2+(y-1)^2=4#

If B is translated by #((-4),(-1))#, then its centre is translated by this.

Centre:

#((8),(5))+((-4),(-1))=((4),(4))#

Circle B:

#(x-4)^2+(y-4)^2=36#

By using the distance between the centres and the radii we can deduce the following:

Let:

#d="distance between centres"#

#r_1,r_2="radii"#

If:

#d > r_1+r_2# Then the circles do not touch.

#d < r_1+r_2# Then the circles intersect at two points or one circle is contained in the other.

#d=r_1+r_2# Then the circles touch at one point.

Using the distance formula:

#d=sqrt((3-4)^2+(1-4)^2)=sqrt(10)#

Sum of radii:

#2+6=8#

#sqrt(10)<8#

So the circles intersect at two points or one is contained in the other. This can be tested by noticing that if the diameter of the smaller circle is less than the radius of the larger then the smaller circle is contained in the larger one.

Diameter of smaller circle is #4#

Radius of larger circle is #6#

So smaller circle is contained in the larger.

To find the shortest distance:

#"radius of B"-("radius of " A+d)#

#6-(2+sqrt(10))=4-sqrt(10)color(white)(888)"units"#

PLOT:

enter image source here

Answer:

#C = (2-sqrt{3},4-2sqrt{3})#

Explanation:

We're oddly given a circle in this problem and awkward wording asking for the smallest possible x coordinate.

I'm not sure what the circle has to do with anything. We could list equations endlessly for circles containing these two points. Forget about the circle.

Given one side, there will be two equilateral triangles with that as a side. Let's rewrite the problem:

Find all possible third vertices #C(x,y)# which make an equilateral triangle with #A(a,b)# and #B(c,d)#.

There are a few different ways to do this. I'd lean toward complex numbers, but I probably should try to keep this simpler than that.

Let's review. The altitude #h# of an equilateral triangle bisects a side #s# so #(s/2)^2 + h^2 = s^2# or # h= \sqrt{3}/2 s#

We just need to go #pm h# along the perpendicular bisector of AB and we'll get to our two possible Cs.

The midpoint D of AB is #D({a+c}/2, {b+d}/2)#

The direction vector from A to B is #B-A=(c-a,d-b)#.

For the perpendicular direction vector we swap and negate one:

#P=(d-b,a-c)#

Now schematically what we're doing is

# C = D \pm h P/|P| = D \pm h/|P| \ P #

We have #s=|AB|=|P|# so # h/|P|= sqrt3/2 #

#C = D pm sqrt{3}/2 P #

# C = ({a+c}/2, {b+d}/2) pm sqrt3/2 (d-b,a-c) #

That's the general solution; let's apply it to

# (a,b)=(0,5), (c,d)=(4,3)#

# C= ( (0+4)/2, (5+3)/2) pm sqrt3/2 (-2, -4) #

# C = (2,4) pm sqrt(3)(1,2) #

# C=(2+sqrt{3},4+2sqrt{3}) or (2-sqrt{3},4-2sqrt{3})#

Uh, the one with the least x coordinate is

#C = (2-sqrt{3},4-2sqrt{3})#

Check:

We check the squared distance to each point

# |C-A|^2 = (2-sqrt{3})^2 + (4-2sqrt{3}-5)^2 = 7 -4sqrt{3} + 13 +4 sqrt{3} = 20#

# |C-B|^2 = (2-sqrt{3}- 4 )^2 + (4-2sqrt{3} - 3 )^2 = 20 quad sqrt#

Answer:

See below.

Explanation:

This question is a little ambiguous. I'm guessing you mean a volume of revolution. If this is the case then a rectangle rotated about a line could be a solid cylinder or a hollow cylinder ( Tube) if the line of rotation is parallel or perpendicular to the sides of the rectangle. If the line of rotation is neither parallel nor perpendicular then a vast number of different solids could result. A big problem here is on the definition. Strictly a prism has a base that is a polygon and faces that are flat, but you could say a cylinder has a polygonal base with an in finite number of sides and an infinite number of faces.

Here are two rotations of a rectangle around different lines.

#y=0#

enter image source here

#x=-2#

enter image source here

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  • Dean R. answered · Yesterday
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