From a point R(5, 8) two tangents RP and RQ are drawn to a given circle s=0 whose radius is 5. If circumcentre of the triangle PQR is (2, 3), then the equation of circle S = 0 is

Options

A. `x^(2)+y^(2)+2x+4y-20=0`
B. `x^(2)+y^(2)+x+2y-10=0`
C. `x^(2)+y^(2)-x-2y-20=0`
D. `x^(2)+y^(2)-4x-6y-12=0`

Explanation

To find the equation of the circle \( S = 0 \) given the point \( R(5, 8) \), the radius of the circle as 5, and the circumcenter of triangle \( PQR \) as \( (2, 3) \), we can follow these steps:

### Step 1: Determine the coordinates of point C
The circumcenter \( (2, 3) \) is the midpoint of segment \( RQ \). Let the coordinates of point \( Q \) be \( (x, y) \). The midpoint formula gives us:

\[ \left( \frac{5 + x}{2}, \frac{8 + y}{2} \right) = (2, 3) \]

From this, we can set up two equations:

1. \( \frac{5 + x}{2} = 2 \)
2. \( \frac{8 + y}{2} = 3 \)

### Step 2: Solve for x and y
From the first equation:

\[ 5 + x = 4 \implies x = 4 - 5 = -1 \]

From the second equation:

\[ 8 + y = 6 \implies y = 6 - 8 = -2 \]

Thus, the coordinates of point \( Q \) are \( (-1, -2) \).

### Step 3: Find the center of the circle
Since \( P \) and \( Q \) are points on the circle and \( R \) is the external point, we can find the center of the circle \( C \) using the circumcenter and point \( R \). The circumcenter \( (2, 3) \) is the midpoint of \( R \) and the center \( C \).

Let the coordinates of the center \( C \) be \( (h, k) \). Using the midpoint formula again:

\[ \left( \frac{5 + h}{2}, \frac{8 + k}{2} \right) = (2, 3) \]

This gives us two equations:

1. \( \frac{5 + h}{2} = 2 \)
2. \( \frac{8 + k}{2} = 3 \)

### Step 4: Solve for h and k
From the first equation:

\[ 5 + h = 4 \implies h = 4 - 5 = -1 \]

From the second equation:

\[ 8 + k = 6 \implies k = 6 - 8 = -2 \]

Thus, the center of the circle \( C \) is \( (-1, -2) \).

### Step 5: Write the equation of the circle
The standard form of the equation of a circle is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Substituting \( h = -1 \), \( k = -2 \), and \( r = 5 \):

\[ (x + 1)^2 + (y + 2)^2 = 5^2 \]

This simplifies to:

\[ (x + 1)^2 + (y + 2)^2 = 25 \]

### Step 6: Expand the equation
Expanding the equation:

\[ (x^2 + 2x + 1) + (y^2 + 4y + 4) = 25 \]

Combining like terms:

\[ x^2 + y^2 + 2x + 4y + 5 - 25 = 0 \]

This simplifies to:

\[ x^2 + y^2 + 2x + 4y - 20 = 0 \]

### Final Answer
The equation of the circle \( S = 0 \) is:

\[ x^2 + y^2 + 2x + 4y - 20 = 0 \]