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

Answer

To find the equation of the circle that touches the line \(2x + 3y + 1 = 0\) at the point \((1, -1)\) and is orthogonal to another circle defined by the diameter endpoints \((0, -1)\) and \((-2, 3)\), we can follow these steps:

### Step 1: Identify the center and radius of the circle touching the line
The general equation of a circle can be written as:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius. Since the circle touches the line at the point \((1, -1)\), we can express the equation of the circle as:
\[ (x - 1)^2 + (y + 1)^2 = r^2 \]
This implies that the center of the circle is \((h, k) = (1, -1)\) and it will be tangent to the line.

### Step 2: Find the distance from the center to the line
The distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(2x + 3y + 1 = 0\) (where \(A = 2\), \(B = 3\), and \(C = 1\)), and the point \((1, -1)\):
\[ d = \frac{|2(1) + 3(-1) + 1|}{\sqrt{2^2 + 3^2}} = \frac{|2 - 3 + 1|}{\sqrt{4 + 9}} = \frac{|0|}{\sqrt{13}} = 0 \]
Since the distance is zero, the radius \(r\) of the circle is equal to the distance from the center to the line, which is \(0\).

### Step 3: Find the equation of the circle with the given diameter
The endpoints of the diameter are \((0, -1)\) and \((-2, 3)\). The center \((h, k)\) of the circle can be found as:
\[ h = \frac{x_1 + x_2}{2} = \frac{0 - 2}{2} = -1 \]
\[ k = \frac{y_1 + y_2}{2} = \frac{-1 + 3}{2} = 1 \]
The radius \(R\) can be calculated as half the distance between the endpoints:
\[ R = \frac{1}{2} \sqrt{(0 + 2)^2 + (-1 - 3)^2} = \frac{1}{2} \sqrt{4 + 16} = \frac{1}{2} \sqrt{20} = \frac{\sqrt{20}}{2} = \sqrt{5} \]
Thus, the equation of the circle with these endpoints as diameter is:
\[ (x + 1)^2 + (y - 1)^2 = 5 \]

### Step 4: Set up the orthogonality condition
For two circles to be orthogonal, the following condition must hold:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Where \(g\) and \(f\) are the coefficients from the general form of the circle equations \(x^2 + y^2 + 2gx + 2fy + c = 0\).

From the first circle:
\[ g_1 = 1, f_1 = -1, c_1 = r^2 \]
From the second circle:
\[ g_2 = -1, f_2 = 1, c_2 = -5 \]

Substituting these values into the orthogonality condition:
\[ 2(1)(-1) + 2(-1)(1) = r^2 - 5 \]
\[ -2 - 2 = r^2 - 5 \]
\[ -4 = r^2 - 5 \implies r^2 = 1 \]

### Step 5: Final equation of the circle
Now substituting \(r^2 = 1\) back into the equation of the first circle:
\[ (x - 1)^2 + (y + 1)^2 = 1 \]
Expanding this gives:
\[ (x^2 - 2x + 1) + (y^2 + 2y + 1) = 1 \]
\[ x^2 + y^2 - 2x + 2y + 1 = 1 \]
\[ x^2 + y^2 - 2x + 2y = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 2x + 2y = 0 \]