Line 2x+3y+1=0 is a tangent to the circle at (1,-1). This circle is orthogonal to a circle which is drawn having diameter as a line segment with end points (0, -1) and (-2, 3). Then the equation of the circle is

Answer

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

Explanation

The equation of circle having tangent `2x + 3y +1 =0` at (1, -1)
`rArr (x-1)^(2)+(y+1)^(2)+lambda(2x+3y+1)=0`
`x^(2)+y^(2)+2x(lambda-1)+y(3lambda+2)+(lambda+2)=0" "...(i)`
which is orthogonal to the circle having end point of diameter (0, 1) and (-2, 3).
`rArrx(x+2)+(y+1)(y-3)=0`
`or x^(2)+y^(2)+2x-2y-3=0" "...(ii)`
`therefore (2(2lambda-2))/(2)*1+ (2(3lambda-2))/(2)(-1)=lambda+2+3`
`rArr 2lambda-2-3lambda-2=lambda-1`
`rArr 2lambda=-3`
`rArr lambda=-3//2`
From Eq(i) equation of circle,
`2x^(2)=2y^(2)-10x-5y+1=0`