Since we know that 3x−y=6 touches the circle at (1,−3) we can form the
structure of the equation of circle.
consider (x−1)2+(y+3)2=0. It just satisfies one point (1,−3) Add to it a
multiple of equation of the line.
(x−1)2+(y+3)2−2λ(3x−y−6)=0. it is a circle of variable radius
which touches the given line at (1,−3).
Expanding
x2+y2−2(1+3λ)+2(3+λ)+10+12λ=0--(1)
It should also touch y=x one can use the concept of perpendicular distance
from center equals radius. But intersection concept is much easier because of
simple equation y=x
By substituting y=xin(1)
x2+2(1−λ)x+5+6λ=0
This must be a perfect since the tangent intersects circle at two identical
points.
so (1−λ)2x+5+6λ=0
This has solution λ=4±2√5
Now we can find the radius
Note R=√g2+f2−c in x2+y2+2gx+2fy+c=0
R2=(1+3λ)2+(3+λ)2−10−12λ
=10λ2
So R=√10|λ|
=√10(4+2√5) or √10(2√5−4)
The smaller of these is
√10(2√5−4)
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