Equation of the circle touching the circle `x^(2) + y^(2) -15x + 5y =0` at ( 1,2) and also passing through the point (0,2) is `:`

Options

A. `x^(2) +y^(2) -x-2y =0`
B. `x^(2) + y^(2)+2x=0`
C. `13( x^(2) +y^(2))-13x -61y +70 =0`
D. None of these

Answer

To find the equation of the circle that touches the circle given by the equation \(x^2 + y^2 - 15x + 5y = 0\) at the point \((1, 2)\) and also passes through the point \((0, 2)\), we can follow these steps:

### Step 1: Rewrite the given circle's equation
The given equation can be rewritten in standard form. We start with:
\[ x^2 + y^2 - 15x + 5y = 0 \]
To convert this into standard form, we complete the square for \(x\) and \(y\).

**Completing the square for \(x\):**
\[ x^2 - 15x = (x - \frac{15}{2})^2 - \left(\frac{15}{2}\right)^2 \]

**Completing the square for \(y\):**
\[ y^2 + 5y = (y + \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2 \]

Putting it all together:
\[ (x - \frac{15}{2})^2 + (y + \frac{5}{2})^2 = \left(\frac{15}{2}\right)^2 + \left(\frac{5}{2}\right)^2 \]

Calculating the right side:
\[ \left(\frac{15}{2}\right)^2 = \frac{225}{4}, \quad \left(\frac{5}{2}\right)^2 = \frac{25}{4} \]
Thus,
\[ \left(\frac{15}{2}\right)^2 + \left(\frac{5}{2}\right)^2 = \frac{225 + 25}{4} = \frac{250}{4} = \frac{125}{2} \]

So the standard form of the circle is:
\[ (x - \frac{15}{2})^2 + (y + \frac{5}{2})^2 = \frac{125}{2} \]

### Step 2: General form of the second circle
Let the equation of the second circle be:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

### Step 3: Use the point (0, 2)
Substituting the point \((0, 2)\) into the equation:
\[ 0^2 + 2^2 + 2g(0) + 2f(2) + c = 0 \]
This simplifies to:
\[ 4 + 4f + c = 0 \quad \text{(Equation 1)} \]

### Step 4: Use the point (1, 2)
Substituting the point \((1, 2)\):
\[ 1^2 + 2^2 + 2g(1) + 2f(2) + c = 0 \]
This simplifies to:
\[ 1 + 4 + 2g + 4f + c = 0 \quad \text{(Equation 2)} \]
Thus:
\[ 5 + 2g + 4f + c = 0 \quad \text{(Equation 2)} \]

### Step 5: Subtract Equation 1 from Equation 2
Subtracting Equation 1 from Equation 2:
\[ (5 + 2g + 4f + c) - (4 + 4f + c) = 0 \]
This simplifies to:
\[ 1 + 2g - 0f = 0 \implies 2g + 1 = 0 \implies g = -\frac{1}{2} \]

### Step 6: Find \(f\) and \(c\)
Substituting \(g = -\frac{1}{2}\) back into Equation 1:
\[ 4 + 4f + c = 0 \implies c = -4 - 4f \quad \text{(Equation 3)} \]

### Step 7: Find the slope of the tangent
To find the slope of the tangent at the point \((1, 2)\) for the first circle, we differentiate:
\[ 2x + 2y \frac{dy}{dx} - 15 + 5 \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx}(2y + 5) = 15 - 2x \]
At the point \((1, 2)\):
\[ \frac{dy}{dx} = \frac{15 - 2(1)}{2(2) + 5} = \frac{13}{9} \]

### Step 8: Use the slope to find \(f\)
The slope of the tangent line for the second circle must be equal to this slope:
\[ \frac{dy}{dx} = \frac{1 - 2f}{2y + 2f} \]
Substituting \((1, 2)\):
\[ \frac{13}{9} = \frac{1 - 2f}{4 + 2f} \]
Cross-multiplying gives:
\[ 13(4 + 2f) = 9(1 - 2f) \]
Expanding and simplifying:
\[ 52 + 26f = 9 - 18f \implies 44f = -43 \implies f = -\frac{43}{44} \]

### Step 9: Substitute \(f\) back to find \(c\)
Substituting \(f\) into Equation 3:
\[ c = -4 - 4(-\frac{43}{44}) = -4 + \frac{172}{44} = -\frac{176}{44} + \frac{172}{44} = -\frac{4}{44} = -\frac{1}{11} \]

### Step 10: Write the final equation
Now substituting \(g\), \(f\), and \(c\) back into the general form:
\[ x^2 + y^2 - x - \frac{43}{22}y - \frac{1}{11} = 0 \]
Multiplying through by 22 to eliminate fractions:
\[ 22x^2 + 22y^2 - 22x - 43y - 2 = 0 \]

### Final Answer
The equation of the circle is:
\[ 22x^2 + 22y^2 - 22x - 43y - 2 = 0 \]