Question

Find the radius of smallest circle touching $3x-y=6$ at point $(1,-3)$and also touches the line $y=x$ ?

Answer 1

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.

$(x-1{)}^2+(y+3{)}^2-2\lambda (3x-y-6)=0$. it is a circle of variable radius

which touches the given line at $(1,-3)$.

Expanding

$x^2+y^2-2(1+3\lambda )+2(3+\lambda )+10+12\lambda =0$--(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 $y=xin\left(1\right)$

$x^2+2(1-\lambda )x+5+6\lambda =0$

This must be a perfect since the tangent intersects circle at two identical

points.

so $(1-\lambda {)}^{2}x+5+6\lambda =0$

This has solution $\lambda =4\pm 2\sqrt{5}$

Now we can find the radius

Note $R=\sqrt{g^2+f^2-c}$ in $x^2+y^2+2gx+2fy+c=0$

$R^2=(1+3\lambda {)}^2+(3+\lambda {)}^2-10-12\lambda$

$=10{\lambda }^2$

So $R=\sqrt{10}\left|\lambda \right|$

$=\sqrt{10}(4+2\sqrt{5})$ or $\sqrt{10}(2\sqrt{5}-4)$

The smaller of these is

$\sqrt{10}(2\sqrt{5}-4)$