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, \[ p + q - r = 2. \]

Answer 2

Given:
- A circle \( C_1: x^2 + y^2 - 4x - 6y - 12 = 0 \)
- Another circle \( C_2: x^2 + y^2 + p x + q y + r = 0 \) with radius 3 units
- \( C_2 \) touches \( C_1 \) internally at point \( (-1, -1) \)

Step 1: Find the center and radius of \( C_1 \).
Rewrite \( C_1 \):
\[ x^2 - 4x + y^2 - 6y = 12 \] Complete the squares:
\[ (x^2 - 4x + 4) + (y^2 - 6y + 9) = 12 + 4 + 9 = 25 \] \[ (x - 2)^2 + (y - 3)^2 = 25 \] So, center of \( C_1 \) is \( O_1 = (2,3) \) and radius \( r_1 = 5 \).

Step 2: Center of \( C_2 \) is \( O_2 = \left(-\frac{p}{2}, -\frac{q}{2}\right) \) and radius \( r_2 = 3 \).

Step 3: Since the two circles touch internally at \( (-1, -1) \), the point lies on both circles:
Substitute \( (-1, -1) \) into \( C_2 \):
\[ (-1)^2 + (-1)^2 + p(-1) + q(-1) + r = 0 \] \[ 1 + 1 - p - q + r = 0 \implies 2 - p - q + r = 0 \] \[ r = p + q - 2 \]

Step 4: The point \( (-1, -1) \) lies on the line joining the centers \( O_1 \) and \( O_2 \). Since the circles touch internally, centers and point of contact are collinear:
\[ \frac{-1 - 2}{x_2 - 2} = \frac{-1 - 3}{y_2 - 3} \] where \( (x_2, y_2) = \left(-\frac{p}{2}, -\frac{q}{2}\right) \). Substitute:
\[ \frac{-3}{-\frac{p}{2} - 2} = \frac{-4}{-\frac{q}{2} - 3} \] Simplify denominators:
\[ \frac{-3}{-\frac{p+4}{2}} = \frac{-4}{-\frac{q+6}{2}} \implies \frac{-3}{-\frac{p+4}{2}} = \frac{-4}{-\frac{q+6}{2}} \] Multiply numerator and denominator:
\[ \frac{-3}{-\frac{p+4}{2}} = \frac{-3 \times 2}{-(p+4)} = \frac{-6}{-(p+4)} = \frac{6}{p+4} \] Similarly:
\[ \frac{-4}{-\frac{q+6}{2}} = \frac{-4 \times 2}{-(q+6)} = \frac{-8}{-(q+6)} = \frac{8}{q+6} \] Equate:
\[ \frac{6}{p+4} = \frac{8}{q+6} \implies 6(q + 6) = 8(p + 4) \] \[ 6q + 36 = 8p + 32 \implies 8p - 6q = 4 \]

Step 5: Distance between centers equals difference of radii (since internally tangent):
\[ |O_1 O_2| = r_1 - r_2 = 5 - 3 = 2 \] Calculate distance:
\[ \sqrt{\left(-\frac{p}{2} - 2\right)^2 + \left(-\frac{q}{2} - 3\right)^2} = 2 \] Square both sides:
\[ \left(-\frac{p}{2} - 2\right)^2 + \left(-\frac{q}{2} - 3\right)^2 = 4 \] Rewrite:
\[ \left(\frac{p + 4}{2}\right)^2 + \left(\frac{q + 6}{2}\right)^2 = 4 \] Multiply both sides by 4:
\[ (p + 4)^2 + (q + 6)^2 = 16 \]

Step 6: From step 4, \( 8p - 6q = 4 \implies 4p - 3q = 2 \).
Express \( q \) in terms of \( p \):
\[ 4p - 3q = 2 \implies 3q = 4p - 2 \implies q = \frac{4p - 2}{3} \]

Step 7: Substitute \( q \) into circle equation:
\[ (p + 4)^2 + \left( \frac{4p - 2}{3} + 6 \right)^2 = 16 \] Simplify inside second bracket:
\[ \frac{4p - 2}{3} + 6 = \frac{4p - 2 + 18}{3} = \frac{4p + 16}{3} \] So:
\[ (p + 4)^2 + \left( \frac{4p + 16}{3} \right)^2 = 16 \] Multiply both sides by 9:
\[ 9 (p + 4)^2 + (4p + 16)^2 = 144 \] Expand:
\[ 9 (p^2 + 8p + 16) + (16 p^2 + 128 p + 256) = 144 \] \[ 9 p^2 + 72 p + 144 + 16 p^2 + 128 p + 256 = 144 \] \[ 25 p^2 + 200 p + 400 = 144 \] \[ 25 p^2 + 200 p + 256 = 0 \] Divide by 25:
\[ p^2 + 8 p + \frac{256}{25} = 0 \] Use quadratic formula:
\[ p = \frac{-8 \pm \sqrt{64 - 4 \times \frac{256}{25}}}{2} = \frac{-8 \pm \sqrt{64 - \frac{1024}{25}}}{2} \] \[ = \frac{-8 \pm \sqrt{\frac{1600 - 1024}{25}}}{2} = \frac{-8 \pm \sqrt{\frac{576}{25}}}{2} = \frac{-8 \pm \frac{24}{5}}{2} \] Two solutions:
\[ p = \frac{-8 + \frac{24}{5}}{2} = \frac{-\frac{40}{5} + \frac{24}{5}}{2} = \frac{-\frac{16}{5}}{2} = -\frac{8}{5} \] \[ p = \frac{-8 - \frac{24}{5}}{2} = \frac{-\frac{40}{5} - \frac{24}{5}}{2} = \frac{-\frac{64}{5}}{2} = -\frac{32}{5} \]

Step 8: Find corresponding \( q \) values:
\[ q = \frac{4p - 2}{3} \] For \( p = -\frac{8}{5} \):
\[ q = \frac{4 \times -\frac{8}{5} - 2}{3} = \frac{-\frac{32}{5} - 2}{3} = \frac{-\frac{32}{5} - \frac{10}{5}}{3} = \frac{-\frac{42}{5}}{3} = -\frac{14}{5} \] For \( p = -\frac{32}{5} \):
\[ q = \frac{4 \times -\frac{32}{5} - 2}{3} = \frac{-\frac{128}{5} - 2}{3} = \frac{-\frac{128}{5} - \frac{10}{5}}{3} = \frac{-\frac{138}{5}}{3} = -\frac{46}{5} \]

Step 9: Calculate \( p + q - r \) with \( r = p + q - 2 \) (from Step 3):
\[ p + q - r = p + q - (p + q - 2) = 2 \]

Therefore,
\[ \boxed{2} \]