The radius of the circle touching the line `2x + 3y +1 = 0` at (1,-1) and cutting orthogonally the circle having line segment joining (0, 3) and (-2,-1) as diameter is

Answer

`2x^(2)+2y^(2)-10x-5y+1=0`

Explanation

We are given that line `2x+3y+1=0` touches the circle `S=0` at `(1,-1)`.

So, the equation of thix circle can be given by
`(x-1)^(2)+(y+1)^(2)+lambda(2x+3y+1)=0,lambda in R`
[Here, `(x-1)^(2)+(y+1)^(2)=0` represents a point circle at (1,-1).]
or `x^(2)+y^(2)+2x(lambda-1)+y(3lambda+2)+(lambda+2)=0` (1)
But his circle is orthogonal to the circle the extremities of whose diameter are (0,3) and (-2,-1) , i.e.,
`x(x+2)+(y-3)(y+1)=0`
or `x^(2)+y^(2)+2x-2y-3=0` (2)
Applying the condition of orthogonality for (1) and (2), we get
`2(lambda-1)xx1+2((3lambda+2)/(2))xx(-1)=lambda+2+(-3)` `[2g_(1)g_(2)+2f_(1)f_(2)=c_(1)+c_(2)]`
or `2 lambda-2-3lambda-2=lambda-1`
or `2 lambda = -3`
or `lambda = -(3)/(2)`
Substituting this value of `lambda `in (1), we get the required circle is
`x^(2)+y^(2)-5x-(5)/(2)y+(1)/(2)=0`
or `2x^(2)+2y^(2)-10x-5y+1=0`