The locus of centre of circle passing through (a, b) and cuts orthogonally the circle `x^(2)+y^(2)=p^(2)` is

Options

A. `2ax+2by-(a^(2)+b^(2)+p^(2))=0`
B. `2ax+2by-(a^(2)-b^(2)+p^(2))=0`
C. `x^(2)+y^(2)-3ax-4by +(a^(2)+b^(2)-p^(2))=0`
D. `x^(2)+y^(2)-2ax+3by+(a^(2)-b^(2)+p^(2))=0`

Answer

To find the locus of the center of a circle that passes through the point (a, b) and cuts orthogonally the circle defined by the equation \(x^2 + y^2 = p^2\), we will follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle can be written as:
\[ x^2 + y^2 + 2Gx + 2Fy + C = 0 \]
where the center of the circle is at the point \((-G, -F)\).

### Step 2: Satisfy the condition for passing through (a, b)
Since the circle passes through the point (a, b), we substitute \(x = a\) and \(y = b\) into the circle equation:
\[ a^2 + b^2 + 2Ga + 2Fb + C = 0 \]
This gives us our first equation (let's call it Equation 1):
\[ a^2 + b^2 + 2Ga + 2Fb + C = 0 \]

### Step 3: Condition for orthogonality with the given circle
The given circle is defined by the equation \(x^2 + y^2 = p^2\). For two circles to intersect orthogonally, the following condition must hold:
\[ 2G_1G_2 + 2F_1F_2 = C_1 + C_2 \]
Here, \(G_1 = G\), \(F_1 = F\), \(C_1 = C\), and for the given circle, \(G_2 = 0\), \(F_2 = 0\), and \(C_2 = -p^2\). Therefore, the orthogonality condition simplifies to:
\[ 0 + 0 = C - p^2 \]
This gives us our second equation (let's call it Equation 2):
\[ C = p^2 \]

### Step 4: Substitute C in Equation 1
Substituting \(C = p^2\) into Equation 1, we get:
\[ a^2 + b^2 + 2Ga + 2Fb + p^2 = 0 \]

### Step 5: Rearranging the equation
Rearranging the above equation gives us:
\[ 2Ga + 2Fb = - (a^2 + b^2 + p^2) \]

### Step 6: Expressing in terms of the center coordinates
We know that the center of the circle is \((-G, -F)\). Therefore, we can express \(G\) and \(F\) in terms of \(X\) and \(Y\) (where \(X = -G\) and \(Y = -F\)):
\[ G = -X, \quad F = -Y \]
Substituting these into the equation gives:
\[ 2(-X)a + 2(-Y)b = - (a^2 + b^2 + p^2) \]
This simplifies to:
\[ -2Xa - 2Yb = - (a^2 + b^2 + p^2) \]
or
\[ 2Xa + 2Yb = a^2 + b^2 + p^2 \]

### Step 7: Final equation for the locus
Dividing through by 2, we arrive at the locus equation:
\[ Xa + Yb = \frac{1}{2}(a^2 + b^2 + p^2) \]

### Step 8: Rearranging to standard form
This can be rearranged to:
\[ 2AX + 2BY - (a^2 + b^2 + p^2) = 0 \]
where \(A = a\) and \(B = b\).

### Conclusion
Thus, the locus of the center of the circle that passes through the point (a, b) and cuts orthogonally the circle \(x^2 + y^2 = p^2\) is given by:
\[ 2AX + 2BY - (a^2 + b^2 + p^2) = 0 \]