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I would be very grateful if you show m the solution of this exercise:

 

*Equation: x^2+y^2-6x-2y+8=0 Find the center and the radius. I found them and i believe the values are for C (3,1) and for R=root(2)

 

*Find the equation of the tangent to the circle above that passes through the beginning of axis O (0,0)

Sorry for my english.

I hope you'll understand it.

Thank you.

Guest Mar 9, 2017

Best Answer 

 #2
avatar+10657 
+5

x^2+y^2-6x-2y+8=0

Completing the squares results in

(x-3)^2 + (y-1)^2 = 2           so the circle center is at    3,1    and the radius is sqrt(2)   as you found

 

tangent line passing through  (0,0) ?   There are TWO of them....

Since the line passes through 0,0     the qaution of the line will be of the form  y= kx      (b will be zero)

Substitute this into the circle equation:

(x-3)^2 + (kx-1)^2 =2  and simplify to :

x^2(1+k^2) + x(6+2k) + 8 =0      (Form:   ax^2 + bx +c =0)

 

we want   b^2 = 4ac    substituting and using the quadratic formula results in k = 1 and -1/7

so your two tangent lines passing through 0,0  will be

    y = x     and y=-1/7x

ElectricPavlov  Mar 9, 2017
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4+0 Answers

 #1
avatar+18777 
+5

*Find the equation of the tangent to the circle above that passes through the beginning of axis O (0,0)

 

1. Center and radius of the circle:

\(\begin{array}{|rcll|} \hline x^2+y^2-6x-2y+8 &=& 0 \\ x^2-6x+y^2-2y &=& -8 \\ (x-\frac{6}{2})^2-3^2 +(y-\frac{2}{2})^2 -1 &=& -8 \\ (x-3)^2 +(y-1)^2 &=& 9+1-8 \\ (x-3)^2 +(y-1)^2 &=& 2 \\\\ Center = C(3,1) && Radius\ r= \sqrt{2} \\ \hline \end{array}\)

 

\(OC = \sqrt{x_c^2+y_c^2} =\sqrt{3^2+1^2}= \sqrt{10}\)

\(\begin{array}{|rcll|} \hline \tan(\alpha) &=& \frac{y_c}{x_c} \\ &=& \frac{1}{3} \\ \tan(\beta) &=& \frac{r}{\sqrt{OC^2-r^2}} \\ &=& \frac{\sqrt{2}}{\sqrt{10-2}} \\ &=& \frac{1}{2} \\ \hline \end{array} \)

 

2. Tangent above slope m:

\(\begin{array}{|rcll|} \hline m=\tan{(\alpha+\beta)} &=& \frac{\tan(\alpha)+\tan(\beta)} {1-\tan(\alpha)\tan(\beta)} \\ m &=& \frac{\frac{1}{3}+\frac{1}{2}} {1-\frac{1}{3}\cdot \frac{1}{2}} \\ m &=& 1 \\\\ y &=& mx +b \quad & | \quad m = 1 \qquad b = 0 \\ \mathbf{y} &\mathbf{=}& \mathbf{x} \\ \hline \end{array} \)

 

3. Tangent below slope m:

\(\begin{array}{|rcll|} \hline m=\tan{(\alpha-\beta)} &=& \frac{\tan(\alpha)+\tan(\beta)} {1+\tan(\alpha)\tan(\beta)} \\ m &=& \frac{\frac{1}{3}-\frac{1}{2}} {1+\frac{1}{3}\cdot \frac{1}{2}} \\ m &=& -\frac{1}{7} \\\\ y &=& mx +b \quad & | \quad m = -\frac{1}{7} \qquad b = 0 \\ \mathbf{y} &\mathbf{=}& \mathbf{-\frac{1}{7}\cdot x} \\ \hline \end{array}\)

 

laugh

heureka  Mar 9, 2017
 #3
avatar+18777 
0

The graph:

 

 

laugh

heureka  Mar 9, 2017
 #2
avatar+10657 
+5
Best Answer

x^2+y^2-6x-2y+8=0

Completing the squares results in

(x-3)^2 + (y-1)^2 = 2           so the circle center is at    3,1    and the radius is sqrt(2)   as you found

 

tangent line passing through  (0,0) ?   There are TWO of them....

Since the line passes through 0,0     the qaution of the line will be of the form  y= kx      (b will be zero)

Substitute this into the circle equation:

(x-3)^2 + (kx-1)^2 =2  and simplify to :

x^2(1+k^2) + x(6+2k) + 8 =0      (Form:   ax^2 + bx +c =0)

 

we want   b^2 = 4ac    substituting and using the quadratic formula results in k = 1 and -1/7

so your two tangent lines passing through 0,0  will be

    y = x     and y=-1/7x

ElectricPavlov  Mar 9, 2017
 #4
avatar+79819 
0

Here's another way to do this.....maybe not as "clean" as heureka's and EP's answers....but....I might see it better....!!!!

 

Borrowing from their answers, the center of the circle is  at (3,1)   and the radius of this circle is

sqrt (2)....so......Call O  =(0,0)   and call (3,1)  = C.....and the distance from  O to C  =

sqrt(3^2 + 1^2)   =  sqrt (10)

 

So....OC  will form the hypotenuse of two right triangles  with both having a leg length = to the raidus of the circle  = sqrt (2)....and.......the remaining sides of these triangles will be the length of two segments drawn from O to the intersection points of the lines with the circle.

 

And this distance is sqrt ( 10 - 2)  = sqrt (8)

 

So....a circle centered at (0,0)  with a radius of sqrt (8)  will intersect the circles at the same points as where the lines intersect the circle

 

So....the equation of the circle will be   x^2 + y^2   = 8 

 

Putting this into  x^2+y^2-6x-2y+8=0   we have that

 

8 - 6x - 2y + 8 = 0

16 - 6x = 2y

8 - 3x   = y

 

Subbing this into  x^2 + y^2  = 8   we have

 

x^2  + (8 - 3x) ^2   = 8

x^2 + 9x^2 - 48x + 64 =  8

10x^2 - 48x + 56 = 0

5x^2 - 24x + 28  = 0    

(5x - 14) (x - 2) = 0   →   x = 14/5 = 2.8  and x  = 2

 

So  when x  = 2  , y = 8 - 3(2)  = 2     and when x = 2.8, y = 8 - 3(2.8)  = -0.4

 

So  the two  intersection points are    (2, 2)   and ( 2.8, -0.4)

 

And since the lines pass through the origin we have

 

y = (2/2)x    →   y  = x        and    y  = (-0.4/ 2.8) →  y  = (-1/7)x

 

Here's a graph showing both lines and both circles :

 

https://www.desmos.com/calculator/xikled50m7

 

 

 

 

 

 

cool cool cool

CPhill  Mar 9, 2017

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