If the point (3,4) lies inside and the point `(-3,-4)` lies outside the circle `x ^(2) + y ^(2) - 7x +5y -p=0,` then the set of all possible values of p is

Options

A. `(24,25)`
B. `(25,26)`
C. `(24,26)`
D. `(0,24)`

Answer

To solve the problem, we need to determine the set of all possible values of \( p \) for the given circle equation:

\[ x^2 + y^2 - 7x + 5y - p = 0 \]

### Step 1: Identify the center and radius of the circle

The general form of a circle is given by:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From the given equation, we can identify:

- \( 2g = -7 \) → \( g = -\frac{7}{2} \)
- \( 2f = 5 \) → \( f = \frac{5}{2} \)
- \( c = -p \)

Thus, the center of the circle is:

\[ \left(-g, -f\right) = \left(\frac{7}{2}, -\frac{5}{2}\right) \]

The radius \( r \) is given by:

\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{7}{2}\right)^2 + \left(\frac{5}{2}\right)^2 + p} \]

Calculating \( g^2 + f^2 \):

\[ g^2 = \left(-\frac{7}{2}\right)^2 = \frac{49}{4}, \quad f^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \]

Thus,

\[ g^2 + f^2 = \frac{49}{4} + \frac{25}{4} = \frac{74}{4} = \frac{37}{2} \]

So, the radius becomes:

\[ r = \sqrt{\frac{37}{2} + p} \]

### Step 2: Use the condition for the point (3, 4) lying inside the circle

For the point \( (3, 4) \) to lie inside the circle, we must have:

\[ (3 - \frac{7}{2})^2 + (4 + \frac{5}{2})^2 < r^2 \]

Calculating the left side:

\[ 3 - \frac{7}{2} = \frac{6}{2} - \frac{7}{2} = -\frac{1}{2} \]
\[ 4 + \frac{5}{2} = \frac{8}{2} + \frac{5}{2} = \frac{13}{2} \]

Now, squaring these values:

\[ \left(-\frac{1}{2}\right)^2 + \left(\frac{13}{2}\right)^2 = \frac{1}{4} + \frac{169}{4} = \frac{170}{4} = \frac{85}{2} \]

Thus, the condition becomes:

\[ \frac{85}{2} < \frac{37}{2} + p \]

Rearranging gives:

\[ p > \frac{85}{2} - \frac{37}{2} = \frac{48}{2} = 24 \]

### Step 3: Use the condition for the point (-3, -4) lying outside the circle

For the point \( (-3, -4) \) to lie outside the circle, we must have:

\[ (-3 - \frac{7}{2})^2 + (-4 + \frac{5}{2})^2 > r^2 \]

Calculating the left side:

\[ -3 - \frac{7}{2} = -\frac{6}{2} - \frac{7}{2} = -\frac{13}{2} \]
\[ -4 + \frac{5}{2} = -\frac{8}{2} + \frac{5}{2} = -\frac{3}{2} \]

Now, squaring these values:

\[ \left(-\frac{13}{2}\right)^2 + \left(-\frac{3}{2}\right)^2 = \frac{169}{4} + \frac{9}{4} = \frac{178}{4} = \frac{89}{2} \]

Thus, the condition becomes:

\[ \frac{89}{2} > \frac{37}{2} + p \]

Rearranging gives:

\[ p < \frac{89}{2} - \frac{37}{2} = \frac{52}{2} = 26 \]

### Step 4: Combine the inequalities

From the two conditions, we have:

\[ 24 < p < 26 \]

### Final Answer

The set of all possible values of \( p \) is:

\[ (24, 26) \]