Find the radius of the smalles circle which touches the straight line `3x-y=6` at `(-,-3)` and also touches the line `y=x` . Compute up to one place of decimal only.

Answer

1.5 unit

Explanation

The equation of the circle touching the line` 3x-y-6=0` at (1,-3) is given by
`(x-1)^(2)+(y+3)^(2)+lambda(3x-y-6)=0`
i.e., `x^(2)+y^(2)+(3lambda-2)x+(6-lambda)y+10-6 lambda=0`
Equation (1) touches `y=x`. Therefore,
`2x^(2)+(2lambda+4)x+10-6 lambda =0`
i.e., `x^(2)+(lambda+2)x+5-3 lambda=0`
has equal roots. Therefore,
`(lambda+2)^(2)-4(5-3 lambda)=lambda^(2)+16lambda-16=0`
or `lambda=(1)/(2)[-16+- sqrt(256+64)]= - 8+- sqrt(80)` ltbr. Now, `("Radius")^(2)=R^(2)=(1)/(4)[(3lambda-2)^(2)+(6-lambda)^(2)-(10-6lambda)4]`
`=(1)/(4)[10lambda^(2)]`
or `R=(sqrt(10)lambda)/(2)=(sqrt(10))/(2)(-8+- 4 sqrt(5))=|-4 sqrt(10)+-10sqrt(2)|`
or Radius of smaller circle `=|-4 sqrt(10)+10sqrt(2)|=1.5` approx.