A circle touches the line $2 x + 3 y + 1 = 0$ at the point $(1, - 1)$ and its orthogonal to the circle which has the line segment having end points $(0, - 1)$ and $(- 2, 3)$ as the diameter.

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Solution

The equation of the circle is
$S_{1} = (x - 1)^{2} + (y + 1)^{2} P t . C i r c l e + λ (2 x + 3 y + 1) T a n g e n t = 0$
$S_{2} =$ The circle on the join of $(0, - 1)$ and $(- 2, 3)$ as diameter i.e., $x (x + 2) + (y + 1) (y - 3) = 0$
The above equations can be re-written as
$S = x^{2} + y^{2} + 2 x (λ - 1) + 2 y (1 + \frac{3 λ}{2}) + (2 + λ) = 0 . . . . . (1)$
$S_{2} = x^{2} + y^{2} + 2 x - 2 y - 3 = 0$
Apply the condition $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$ for orthogonal intersection of $S_{1}$ and $S_{2}$ .
$2 (λ - 1) . 1 + 2 (1 + \frac{3 λ}{2}) (- 1) = 2 + λ - 3$
or $- λ - 4 = λ - 1$ or $- 3 = 2 λ$
or $λ = - 3 / 2$
Putting the value of $λ$ in $(1)$ , the required circle is $2 x^{2} + 2 y^{2} - 10 x - 5 y + 1 = 0$ .

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A circle touches the line 2x+3y+1=0 at the point (1,−1) and its orthogonal to the circle which has the line segment having end points (0,−1) and (−2,3) as the diameter.

A circle touches the line $2 x + 3 y + 1 = 0$ at the point $(1, - 1)$ and its orthogonal to the circle which has the line segment having end points $(0, - 1)$ and $(- 2, 3)$ as the diameter.