The circle through the two points (-2,5),(0,0) and intersecting `x^2+y^2-4x+3y-1=0` orthogonally is

Options

A. `2x^2+2y^2-11x-16y=0`
B. `x^2+y^2-4x-4y=0`
C. `x^2+y^2+2x-5y=0`
D. `2x^2+2y^2+2x-5y+1=0`

Answer

To solve the problem of finding the equation of the circle passing through the points (-2, 5) and (0, 0) that intersects the given circle \(x^2 + y^2 - 4x + 3y - 1 = 0\) orthogonally, we can follow these steps:

### Step 1: General Equation of Circle
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \(g\), \(f\), and \(c\) are constants.

### Step 2: Substitute Point (0, 0)
Since the circle passes through the point (0, 0), we substitute this point into the general equation:
\[ 0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \]
This simplifies to:
\[ c = 0 \]
Thus, the equation of the circle becomes:
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

### Step 3: Substitute Point (-2, 5)
Next, we substitute the point (-2, 5) into the equation:
\[ (-2)^2 + (5)^2 + 2g(-2) + 2f(5) = 0 \]
Calculating this gives:
\[ 4 + 25 - 4g + 10f = 0 \]
This simplifies to:
\[ -4g + 10f = -29 \quad \text{(Equation 1)} \]

### Step 4: Orthogonality Condition
For the circles to intersect orthogonally, the following condition must hold:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Here, for the given circle \(x^2 + y^2 - 4x + 3y - 1 = 0\):
- \(g_2 = -2\) (half of the coefficient of \(x\))
- \(f_2 = \frac{3}{2}\) (half of the coefficient of \(y\))
- \(c_2 = -1\)

Substituting these values into the orthogonality condition gives:
\[ 2g(-2) + 2f\left(\frac{3}{2}\right) = 0 - 1 \]
This simplifies to:
\[ -4g + 3f = -1 \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \(-4g + 10f = -29\)
2. \(-4g + 3f = -1\)

Subtract Equation 2 from Equation 1:
\[ (-4g + 10f) - (-4g + 3f) = -29 + 1 \]
This simplifies to:
\[ 7f = -28 \]
Thus, we find:
\[ f = -4 \]

### Step 6: Substitute \(f\) Back to Find \(g\)
Now substitute \(f = -4\) back into Equation 1:
\[ -4g + 10(-4) = -29 \]
This simplifies to:
\[ -4g - 40 = -29 \]
Thus:
\[ -4g = 11 \quad \Rightarrow \quad g = -\frac{11}{4} \]

### Step 7: Write the Final Equation of the Circle
Now substituting \(g\) and \(f\) back into the general equation:
\[ x^2 + y^2 + 2\left(-\frac{11}{4}\right)x + 2(-4)y = 0 \]
This simplifies to:
\[ x^2 + y^2 - \frac{11}{2}x - 8y = 0 \]
Multiplying through by 2 to eliminate the fraction gives:
\[ 2x^2 + 2y^2 - 11x - 16y = 0 \]

### Final Answer
Thus, the required equation of the circle is:
\[ 2x^2 + 2y^2 - 11x - 16y = 0 \]