Consider the circle `x^(2)+y^(2)_4x-2y+c=0` whose centre is A(2,1). If the point P(10,7) is such that the line segment PA meets the circle in Q with PQ=5 then c=

Options

A. `-15`
B. 20
C. 30
D. `-20`

Answer

To solve the problem, we need to find the value of the constant \( c \) in the equation of the circle given by:

\[ x^2 + y^2 - 4x - 2y + c = 0 \]

We know that the center of the circle is at point \( A(2, 1) \) and that the point \( P(10, 7) \) lies outside the circle such that the line segment \( PA \) meets the circle at point \( Q \) with \( PQ = 5 \).

### Step 1: Calculate the distance \( PA \)

Using the distance formula:

\[ PA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting \( A(2, 1) \) and \( P(10, 7) \):

\[ PA = \sqrt{(10 - 2)^2 + (7 - 1)^2} \]

Calculating the squares:

\[ = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]

### Step 2: Relate distances \( PA \), \( AQ \), and \( PQ \)

From the problem, we know:

\[ PA = AQ + PQ \]

Given \( PQ = 5 \):

\[ 10 = AQ + 5 \]

Thus, we can find \( AQ \):

\[ AQ = 10 - 5 = 5 \]

### Step 3: Identify the radius of the circle

Since \( AQ \) is the distance from the center \( A \) to the point \( Q \) on the circle, we have:

\[ r = AQ = 5 \]

### Step 4: Use the formula for the radius of the circle

The radius \( r \) can also be expressed in terms of the coefficients of the circle's equation:

\[ r = \sqrt{g^2 + f^2 - c} \]

where the standard form of the circle is \( (x - h)^2 + (y - k)^2 = r^2 \) and \( g = -h \), \( f = -k \).

From the equation \( x^2 + y^2 - 4x - 2y + c = 0 \):

- \( g = -\frac{-4}{2} = -2 \)
- \( f = -\frac{-2}{2} = -1 \)

### Step 5: Substitute values into the radius formula

Now substituting \( g \) and \( f \) into the radius formula:

\[ r = \sqrt{(-2)^2 + (-1)^2 - c} \]

Substituting \( r = 5 \):

\[ 5 = \sqrt{4 + 1 - c} \]

### Step 6: Square both sides to eliminate the square root

Squaring both sides gives:

\[ 25 = 5 - c \]

### Step 7: Solve for \( c \)

Rearranging the equation:

\[ c = 5 - 25 \]
\[ c = -20 \]

Thus, the value of \( c \) is:

\[ \boxed{-20} \]