Question

If the equation of the circle whose radius is 3 units and which touches internally the circle $$ x^2 + y^2 - 4x - 6y - 12 = 0 $$ at the point $(-1, -1)$ is $$ x^2 + y^2 + px + qy + r = 0, $$ then $p + q - r$ is: 

• A. \( 2 \)
• B. \( \frac{5}{2} \)
• C. \( \frac{26}{5} \)
• D. \( 3 \)

Hint:

For circles touching internally, use the distance formula between centers and the radii relation \( |R - r| = d \) to solve problems.

Answer 1

The Correct Option is A

Step 1: Finding the center and radius of the given circle The given circle equation: \[ x^2 + y^2 - 4x - 6y - 12 = 0 \] Rewriting in standard form, complete the square: \[ (x - 2)^2 - 4 + (y - 3)^2 - 9 - 12 = 0 \] \[ (x - 2)^2 + (y - 3)^2 = 25 \] Thus, the center is \( (2,3) \) and radius \( R = 5 \). 
Step 2: Finding the required circle The required circle has radius \( r = 3 \) and is internally tangent at \( (-1,-1) \). Using the equation transformation method and substituting \( (-1,-1) \), we determine \( p, q, r \). After calculation, \[ p + q - r = 2. \]