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What is the equation of a circle touching the lines x-3y-11=0 and 3x-y-9=0 and having its center on the line x+2y+19=0?

 

To the one who can solve this please kindly elaborate further so that I may understand it clearly..  Thank you and God Bless.

Guest Dec 13, 2015

Best Answer 

 #3
avatar+91051 
+10

What is the equation of a circle touching the lines x-3y-11=0 and 3x-y-9=0 and having its center on the line x+2y+19=0?

 

Well if the lines just touch the circle then they are tangents.  And the radius is at right angles to the tangent that it touches.

so

 

Let  the centre of the cirlce be (h,k)

 

Perpendicular distance from ax+by+c=0 to (x1,y1) is

 

\(\boxed{d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}}\\~\\ \)

so 

 

x-3y-11=0                                         3x-y-9=0                                            

a=1, b=-3, c=-11        x1=h     y1=k

a=3, b=-1, c=-9     x1=h, y1=k
\(d=\frac{|h-3k-11|}{\sqrt{1+9}}\\~\\ d=\frac{|h-3k-11|}{\sqrt{10}}\\~\\ \)\(d=\frac{|3h-k-9|}{{\sqrt{10}}}\\~\\ \)

Added at end

h=17, k=-18

d= 6sqrt(10)

so the formula would be

\((x-17)^2+(y+18)^2=360\)

 

OR

h=-3,  k=-8

d= sqrt(10)

so the formula would be

\((x+3)^2+(y+8)^2=10\)

Added at end

 

 

Now d is the radius so 

|h-3k-11|=|3h-k-9|

h-3k-11=3h-k-9       or      h-3k-11 = -(3h-k-9)

h-3k = 3h-k+2           or       h-3k-11 = -3h+k+9

-3k = 2h-k+2           or       4h-3k-11 = +k+9

-3k = 2h-k+2           or       4h-3k = +k+20

-2k = 2h +2             or           -4k =  -4h+20

k   =  -h  -1             or            k = h - 5

k   =  -h  -1             or            k = h -5

centre

(h,   -h-1)               or            (h, h-5)

 

Now sub these points into the equation x+2y+19=0

1)     (h,   -(h+1) )

h-2(h+1)+19=0

h-2h-2+19=0

-h+17=0

h=17

 

h= 17,   k= -(17+1) = -18       ( 17, -18)

 

2)    (h, h-5)

h+2(h-5)+19=0

h+2h-10+19 = 0

3h +9 = 0

3h = -9

h= -3                         

k= -3-5 = -8       (-3, -8)

 

I thought I had finished . 

 

I have finished it in the table above

 

The circles are 

 

\((x+3)^2+(y+8)^2=10\\~\\ and\\~\\ (x-17)^2+(y+18)^2=360 \)

 

 

Melody  Dec 13, 2015
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8+0 Answers

 #1
avatar
0

Not  quite sure on that one... Google or other online websites are also great recourses. But I guess depending upon the grade your in you want it to be fairly easy or highly complicated.

Guest Dec 13, 2015
 #2
avatar
0

I think my computer just glitchd. Instead of asking me if I was a robot it asked me if I was a tellituby. Weird LOL

Guest Dec 13, 2015
 #3
avatar+91051 
+10
Best Answer

What is the equation of a circle touching the lines x-3y-11=0 and 3x-y-9=0 and having its center on the line x+2y+19=0?

 

Well if the lines just touch the circle then they are tangents.  And the radius is at right angles to the tangent that it touches.

so

 

Let  the centre of the cirlce be (h,k)

 

Perpendicular distance from ax+by+c=0 to (x1,y1) is

 

\(\boxed{d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}}\\~\\ \)

so 

 

x-3y-11=0                                         3x-y-9=0                                            

a=1, b=-3, c=-11        x1=h     y1=k

a=3, b=-1, c=-9     x1=h, y1=k
\(d=\frac{|h-3k-11|}{\sqrt{1+9}}\\~\\ d=\frac{|h-3k-11|}{\sqrt{10}}\\~\\ \)\(d=\frac{|3h-k-9|}{{\sqrt{10}}}\\~\\ \)

Added at end

h=17, k=-18

d= 6sqrt(10)

so the formula would be

\((x-17)^2+(y+18)^2=360\)

 

OR

h=-3,  k=-8

d= sqrt(10)

so the formula would be

\((x+3)^2+(y+8)^2=10\)

Added at end

 

 

Now d is the radius so 

|h-3k-11|=|3h-k-9|

h-3k-11=3h-k-9       or      h-3k-11 = -(3h-k-9)

h-3k = 3h-k+2           or       h-3k-11 = -3h+k+9

-3k = 2h-k+2           or       4h-3k-11 = +k+9

-3k = 2h-k+2           or       4h-3k = +k+20

-2k = 2h +2             or           -4k =  -4h+20

k   =  -h  -1             or            k = h - 5

k   =  -h  -1             or            k = h -5

centre

(h,   -h-1)               or            (h, h-5)

 

Now sub these points into the equation x+2y+19=0

1)     (h,   -(h+1) )

h-2(h+1)+19=0

h-2h-2+19=0

-h+17=0

h=17

 

h= 17,   k= -(17+1) = -18       ( 17, -18)

 

2)    (h, h-5)

h+2(h-5)+19=0

h+2h-10+19 = 0

3h +9 = 0

3h = -9

h= -3                         

k= -3-5 = -8       (-3, -8)

 

I thought I had finished . 

 

I have finished it in the table above

 

The circles are 

 

\((x+3)^2+(y+8)^2=10\\~\\ and\\~\\ (x-17)^2+(y+18)^2=360 \)

 

 

Melody  Dec 13, 2015
 #4
avatar+78755 
+5

Very impressive, Melody.....I like this one.....!!!!

 

 

cool cool cool

CPhill  Dec 14, 2015
 #5
avatar+4 
+5

The problem only wants the equation of a circle, it means just 1 equation. Your first equation is right but I'm not sure about your second equation answer.

Psycho  Dec 17, 2015
 #6
avatar+26329 
+5

Both of the solutions given by Melody are correct.  Here is an alternative derivation:

 

circle tangents

 

These values for h and k can be put back into the earlier equations to get the corresponding radii.

.

Alan  Dec 17, 2015
edited by Alan  Dec 17, 2015
 #7
avatar+91051 
0

That is really neat Alan  laugh

Your second line where you differentiated the circle. 

A circle is not a function.  I didn't think that you could differentiate it ??      frown

Melody  Dec 17, 2015
 #8
avatar+91051 
0

Thanks Chris.  

And Psycho, I have decided to take your response as a thankyou.  :))

Melody  Dec 17, 2015

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