Question

The equation of the circle whose radius is 3 and which touches the circle \(x^2 + y^2 - 4x - 6y - 12 = 0\) internally at \((-1, -1)\) is

• A. \(5x^2 + 5y^2 - 8x - 14y - 32 = 0\)
• B. \(x^2 + y^2 - 12x - 14y - 28 = 0\)
• C. \(3x^2 + 3y^2 - 8x - 14y - 31 = 0\)
• D. \(x^2 + y^2 - 5x - 7y - 14 = 0\)

Hint:

When a circle touches another internally, the center of the smaller circle lies on the line joining the center of the larger circle and the point of contact, at a distance equal to its radius from the point of contact.

Answer 1

The Correct Option is A

Step 1: Analyze the Given Circle (S1):
Equation: \(x^2 + y^2 - 4x - 6y - 12 = 0\). Center \(C_1 = (2, 3)\). Radius \(R_1 = \sqrt{2^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5\).
Step 2: Determine Position of New Center (\(C_2\)):
The new circle (S2) has radius \(R_2 = 3\) and touches S1 internally at \(P(-1, -1)\). Since \(R_2<R_1\) (3<5), S2 is inside S1. The centers \(C_1, C_2\) and the point of contact \(P\) are collinear. Since the touch is internal, \(C_2\) lies on the segment \(PC_1\). Distance \(PC_1 = R_1 = 5\). Distance \(PC_2 = R_2 = 3\). So \(C_2\) divides \(PC_1\) in the ratio \(3 : (5-3) = 3:2\).
Step 3: Calculate \(C_2\) Coordinates:
Using section formula for \(C_2\) dividing \(P(-1, -1)\) and \(C_1(2, 3)\) in ratio \(3:2\): \[ x_{C2} = \frac{3(2) + 2(-1)}{3+2} = \frac{6 - 2}{5} = \frac{4}{5} \] \[ y_{C2} = \frac{3(3) + 2(-1)}{3+2} = \frac{9 - 2}{5} = \frac{7}{5} \] Center \(C_2 = (\frac{4}{5}, \frac{7}{5})\).
Step 4: Equation of New Circle:
\[ (x - \frac{4}{5})^2 + (y - \frac{7}{5})^2 = 3^2 \] \[ x^2 - \frac{8}{5}x + \frac{16}{25} + y^2 - \frac{14}{5}y + \frac{49}{25} = 9 \] Multiply by 25 to clear denominators: \[ 25x^2 - 40x + 16 + 25y^2 - 70y + 49 = 225 \] \[ 25x^2 + 25y^2 - 40x - 70y + 65 - 225 = 0 \] \[ 25x^2 + 25y^2 - 40x - 70y - 160 = 0 \] Divide by 5: \[ 5x^2 + 5y^2 - 8x - 14y - 32 = 0 \]