+118725 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
\((x+3)^2+(y+8)^2=10\\~\\ and\\~\\ (x-17)^2+(y+18)^2=360 \)
