Question

Find the equation of the circle that touches the line 2x + 3y +1 = 0 at the point (1,-1) and is orthogonal to the circle which has the line segment having end points (0,-1) and (-2,3) as the diameter.


✔️✔️ Proper solution needed ✔️✔️​

Answer 1

Heya mate
The answer of ur question is







 The equation of the circle having tangent 2x + 3y + 1 = 0 at the point (1, -1) is

(x - 1)2 + (y + 1)2 + μ(2x + 3y + 1) = 0
So
x2 + y2 + 2x(μ -1) + y(3μ + 2) + (μ+2) = 0      ….. (1)

As we know that this is orthogonal to the circle having end points of diameter as (0, -1) and (-2, 3)

It would be x(x+2) + (y+1)(y-3) = 0

And also it is , x2 + y2 + 2x -2y -3 = 0

Therefore, [2(2μ – 2)/ 2]. 1 + [2(3μ + 2)/ 2]. (-1) = μ + 2 – 3
2μ - 2 - 3μ – 2= μ -1
2μ = -3
μ = -3/2


Putting the value of μ= - 3/2 In the equation ( 1 ) ,we should get



2x2 + 2y2 -10x -5y + 1 = 0.




Hence equation is finded




hope it helps