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.
+118608 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
\((x+3)^2+(y+8)^2=10\\~\\ and\\~\\ (x-17)^2+(y+18)^2=360 \)
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.
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
+118608 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
\((x+3)^2+(y+8)^2=10\\~\\ and\\~\\ (x-17)^2+(y+18)^2=360 \)
+4 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.
